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Q.1. Suppose that the normals drawn at three different points on the parabola y^{2} = 4x pass through the point (h, k). Show that h > 2. (1981  4 Marks)
Ans. 2
Sol. The equation of a normal to the parabola y^{2} = 4ax in its slope form is given by
y = mx – 2am – am^{3}
∴Eq. of normal to y^{2 }= 4x, is y = mx – 2m – m^{3} ...(1)
Since the normal drawn at three different points on the parabola pass through (h, k), it must satisfy the equation (1)
∴k = mh – 2m – m^{3} ⇒ m^{3} – (h – 2) m + k = 0
This cubic eq. in m has three different roots say m_{1}, m_{2}, m_{3}
∴m_{1} + m_{2} + m_{3} = 0 ...(2)
m_{1}m_{2} + m_{2}m_{3} + m_{3}m_{1 }= – (h – 2) ...(3)
Now, (m_{1} + m_{2} + m_{3})^{2} = 0 [Squaring (2)]
⇒ m_{1}^{2} + m_{2}^{2}+m_{3}^{2} = – 2 (m_{1}m_{2 }+ m_{2}m_{3} + m_{3}m_{1}) ⇒ m_{1}^{2} + m_{2}^{2 }+ m_{3}^{2} = 2 (h – 2) [Using (3)]
Since LHS of this equation is he sum of perfect squares, therefore it is + ve
∴h – 2 > 0 ⇒ h > 2 Proved
Q.2. A is a point on the parabola y^{2} = 4ax. The normal at A cuts the parabola again at point B. If AB subtends a right angle at the vertex of the parabola. find the slope of AB. (1982  5 Marks)
Ans. m =±
Sol. Parabola y^{2 }= 4ax.
Let at any pt A equation of normal is y = mx – 2am – am^{3}. ...(1)
Combined equation of OA and OB can be obtained by making equation of parabola homogeneous with the help of normal.
∴Combined eq. of OA and OB is
[ From eqn. (1) using]
⇒ 4mx^{2} – 4xy – (2m + m^{3}) y^{2} = 0 But angle between the lines represented by this pair is 90°.
⇒ coeff. of x^{2} + coeff of y^{2} = 0 ⇒ 4m – 2m – m^{3} = 0
⇒ m^{3} – 2m = 0 ⇒ m = 0, , –
But for m = 0 eq. of normal becomes y = 0 which does not intersect the parabola at any other point.
∴ m =±
Q.3. Three normals are drawn from the point (c, 0) to the curve y^{2 }= x. Show that c must be greater than 1/2. One normal is always the xaxis. Find c for which the other two normals are perpendicular to each other. (1991  4 Marks)
Ans. c = 3/4
Sol. Given parabola is y^{2 }= x.
Normal is y =
As per question this normal passes through (c, 0) therefore, we get
...(1)
m = 0 shows normal is y = 0 i.e. xaxis is always a normal.
At c = from (1) m = 0
∴for other real values of m, c > 1/2
Now for other two normals to be perpendicular to each other, we must have m_{1}.m_{2} = – 1
Or in other words, if m_{1},m_{2} are roots of = 0, then product of roots = – 1
⇒ c = 3/4
Q.4. Through the vertex O of parabola y^{2} = 4x, chords OP and OQ are drawn at right angles to one another. Show that for all positions of P, PQ cuts the axis of the parabola at a fixed point. Also find the locus of the middle point of PQ. (1994  4 Marks)
Ans. y^{2} = 2 (x – 4)
Sol. Let the equation of chord OP be y = mx.
Then eqn of chord OQ will be [∵ OQ ⊥ OP]
P is pt. of intersection of y = mx and y^{2} = 4x.
Solving the two we get
Q is pt. of intersection of and y^{2 }= 4x.
Solving the two we get Q (4m^{2}, – 4m)
Now eq. of PQ is
⇒ (1 – m^{2}) y + 4m – 4m^{3} = mx – 4m^{3} ⇒ mx – (1 – m^{2}) y – 4m = 0
This line meets xaxis where y = 0 i.e. x = 4
⇒ OL = 4, which is constant as independent of m. Again let (h, k) be the mid pt. of PQ, then
⇒ 2h = k^{2} + 8 ⇒ k^{2} = 2 (h – 4)
∴ Locus of (h, k) is y^{2} = 2 (x – 4)
Q.5. Show that the locus of a point that divides a chord of slope 2 of the parabola y^{2} = 4x internally in the ratio 1: 2 is a parabola. Find the vertex of this parabola. (1995  5 Marks)
Ans. (2/9, 8/9)
Sol. Let P (t_{1}^{2} , 2t_{1}) and Q (t_{2}^{2} , 2t_{2}) be the ends of the chord PQ of the parabola
y^{2 }= 4x ...(1)
∴Slope of chord PQ
⇒ t_{2} + t_{1} = 1 ...(2)
If R(x_{1}, y_{1}) is a point dividing PQ internally in the ratio 1 : 2, then
⇒ t_{2}^{2} + 2t_{1}^{2 }=3x_{1 } ...(3)
and t_{2 }+ 2t_{1}= (3y_{1}) /2 ...(4)
From (2) and (4), we get
Substituting in (3), we get
⇒ (9/4)y_{1}^{2} – 4y_{1} =x_{1}2
∴Locus of the point R ( x_{1} ,y_{1}) is (y – 8/9)^{2} = (4/9) (x – 2/9) which is a parabola having vertex at the point (2/9, 8/9).
Q.6. Let ‘d’ be the perpendicular distance from the centre of the ellipse to the tangent drawn at a point P on the ellipse. If F_{1} and F_{2} are the two foci of the ellipse, then show that
(1995  5 Marks)
Ans.
Sol. Equation to the tangent at the point P (a cosθ, b sinθ) on x^{2}/a^{2} + y^{2}/b^{2} = 1 is
...(1)
∴d = perpendicular distance of (1) from the centre (0, 0) of the ellipse
=
= 4 (a^{2} – b^{2}) cos^{2}θ = 4a^{2 }e^{2 }cos^{2}θ....(2)
The coordinates of focii F_{1} and F_{2} are F_{1} = (ae, 0) and F_{2} = (–ae, 0)
= a (1 – e cosθ) Similarly, PF_{2} = a (1 + e cosθ)
∴(PF_{1 }– PF_{2})^{2} = 4a^{2} e^{2} cos^{2}θ ...(3)
Hence from (2) and (3), we have
(PF_{1} – PF_{2})^{2} =
Q.7. Points A, B and C lie on the parabola y^{2} = 4ax. The tangents to the parabola at A, B and C, taken in pairs, intersect at points P, Q and R. Determine the ratio of the areas of the triangles ABC and PQR. (1996  3 Marks)
Ans. 2 : 1
Sol. Let the three points on the parabola y^{2} = 4ax be A (at_{1}^{2} , 2at_{1}) , B(at_{2}^{2} , 2at_{2}) and C (at_{3}^{2} , 2at_{3}) .
Then using the fact that equation of tangent to y^{2 }= 4ax at (at^{2}, 2at) is , we get equations of tangents at A, B and C as folllows
...(1)
...(2)
...(3)
Solving the above equations pair wise we get the pts.
P (at_{1}t_{2}, a (t_{1} + t_{2}))
Q (at_{2}t_{3}, a (t_{2} + t_{3}))
R (at_{3}t_{1}, a (t_{3} + t_{1}))
Now, area of ΔABC =
...(4)
Also area of ΔPQR
Expanding along C_{1},
...(5)
From equations (4) and (5), we get
∴The required ratio is 2 : 1
Q.8. From a point A common tangents are drawn to the circle x^{2} + y^{2} = a^{2}/2 and parabola y^{2} = 4ax. Find the area of the quadrilateral formed by the common tangents, the chord of contact of the circle and the chord of contact of the parabola. (1996  2 Marks)
Ans.
Sol. This line will touch the circle x^{2} + y^{2} = a^{2}/2
if
⇒ 2 = m^{4} + m^{2 }⇒ m^{4 }+ m^{2} – 2 = 0
⇒ (m^{2} + 2) (m2 – 1) = 0 ⇒ m = 1, – 1
Thus the two tangents (common one) are y = x + a and y = – x – a
These two intersect each other at (– a, 0) The chord of contact at A (– a, 0) for the circle x^{2} + y^{2} = a^{2}/2 is
(– a. x) + 0.y = a^{2}/2 i.e., x = – a/2 and the chord of contact at A (– a,0) for the parabola y^{2} = 4ax is 0.y = 2a (x – a) i.e., x = a
Note that DE is latus recturn of parabola y^{2 }= 4ax, therefore its lengths is 4a.
Chords of contact are clearly parallel to each other, so req. quadrilateral is a trapezium.
Ar (trap BCDE) = BC + DEx KL
Q.9. A tangent to the ellipse x^{2} + 4y^{2} = 4 meets the ellipse x^{2 }+ 2y^{2} = 6 at P and Q. Prove that the tangents at P and Q of the ellipse x^{2 }+ 2y^{2} = 6 are at right angles.(1997  5 Marks)
Ans.
Sol. The given ellipses are
...(1)
and ...(2)
Then the equation of tangent to (1) at any point T (2 cosθ, sinθ) is given by
or ...(3)
Let this tangent meet the ellipse (2) at P and Q.
Let the tangents drawn to ellipse (2) at P and Q meet each other at R (x_{1}, y_{1})
Then PQ is chord of contact of ellipse (2) with respect to the pt R (x_{1}, y_{1}) and is given by
...(4)
Clearly equations (3) and (4) represent the same lines and hence should be identical. Therefore comparing the cofficients, we get
⇒ x_{1} = 3 cosθ , y_{1 }= 3 sinθ ⇒ x_{1}^{2} +y_{1}^{2}= 9
⇒ Locus of (x_{1}, y_{1}) is x^{2} + y^{2} = 9
which is the director circle of the ellipse = and
Thus tangents at P and Q are at right ∠’s.
KEY CONCEPT : We know that the director circle is the locus of intersection point of the tangents which are at right ∠.
Q.10. The angle between a pair of tangents drawn from a point P to the parabola y^{2} =4ax is 45°. Show that the locus of the point P is a hyperbola. (1998  8 Marks)
Ans.
Sol. Let P (e, f) be any point on the locus. Equation of pair of tangents from P (e, f) to the parabola y^{2 }= 4ax is [ fy – 2a (x + e)]^{2 }
= (f ^{2} – 4ae) (y^{2} – 4ax) [T^{2} = SS_{1}]
Here, a = coefficient of x^{2} = 4a^{2} ...(1)
2h = coefficient of xy = – 4af ...(2)
and b = coefficient of y^{2 }= f^{ 2} – (f ^{2} – 4ae) = 4ae ...(3)
If they include an angle 45°, then
1 tan 45°
or,(a + b)^{2} = 4 (h^{2 }– ab)
or, (4a^{2} + 4ae)^{2} = 4 [4a^{2}f ^{2} – (4a^{2}) (4ae)]
or,(a + e)^{2 }= f^{2 }– 4ae or e^{2 }+ 6ae + a^{2} – f^{2} = 0
or(e + 3a)^{2} – f^{ 2} = 8a^{2 }
Hence the required locus is (x + 3a)^{2} – y^{2 }= 8a^{2}, which is a hyperbola.
Q.11. Consider the family of circles x^{2} + y^{2} = r^{2}, 2 < r < 5. If in the first quadrant, the common taingent to a circle of this family and the ellipse 4x^{2} + 25y^{2} = 100 meets the coordinate axes at A and B, then find the equation of the locus of the midpoint of AB. (1999  10 Marks)
Ans. Sol. Let any point P on ellipse 4x^{2} + 25y^{2} = 100 be (5 cosθ, 2 sinθ). So equation of tangent to the ellipse at P will be
Tangent (1) also touches the circle x^{2} + y^{2} = r^{2}, so its distance from origin must be r.
Tangent (2) intersects the coordinate axes at and respectively. Let M (h, k) be the midpoint of line segment AB. Then by mid point formula
⇒
⇒
Hence locus of M (h, k) is
Locus is independent of r.
Q.12. Find the coordinates of all the points P on the ellipse , for which the area of the triangle PON is maximum, where O denotes the origin and N, the foot of the perpendicular from O to the tangent at P.(1999  10 Marks)
Ans.
Sol. The ellipse is ..(1)
Since this ellipse is symmetrical in all four quadrants, either there exists no such P or four points, one in each qudrant.
Without loss of generality we can assume that a > b and P lies in first quadrant.
Let P (a cosθ, b sinθ) then equation of tangent is
Equation of ON is
Equation of normal at P is ax secθ – by cosecθ = a–^{2 } b^{2}
and NP = OL
∴Z = Area of OPN =x ON x NP
Let u= a^{2} tanθ + b^{2} cotθ
⇒ tanθ = b/a
u is minimum at θ = tan^{–1} b/a
So Z is maximum at θ = tan^{–1} b/a
By symmetry, we have four such points
Q.13. Let ABC be an equilateral triangle inscribed in the circle x^{2} + y^{2} = a^{2}. Suppose perpendiculars from A, B, C to the major axis of the ellipse
meets the ellipse respectively, at P, Q, R. so that P, Q, R lie on the same side of the major axis as A, B, C respectively. Prove that the normals to the ellipse drawn at the points P, Q and R are concurrent. (2000  7 Marks)
Ans.
Sol. Let A, B, C be the point on circle whose coordinates are A = [a cosθ , a sinθ ]
and
Further, P [ a cosθ , b sinθ ] (Given)
and
It is given that P, Q, R are on the same side of xaxis as A, B, C.
So required normals to the ellipse are ax secθ– by cosecθ= a^{2} – b^{2} ...(1)
...(2)
...(3)
Now, above three normals are concurrent ⇒ Δ = 0
where Δ
Multiplying and dividing the different rows R_{1}, R_{2 }and R_{3} by sinθ cosθ ,
and respectively, we get
[Operating R_{2} → R_{2} + R_{3} and simplyfing R_{2} we get R_{2} ≡ R_{1}]
Q.14. Let C_{1} and C_{2} be respectively, the parabolas x^{2} = y – 1 and y^{2} = x – 1. Let P be any point on C_{1} and Q be any point on C_{2}. Let P_{1 }and Q_{1 }be the reflections of P and Q, respectively, with respect to the line y = x. Prove that P_{1 }lies on C_{2}, Q_{1} lies on C_{1} an d PQ ≥ min{PP_{1}, QQ_{1}} . Hence or otherwise determine points P_{0} and Q_{0} on the parabolas C_{1} and C_{2} respectively such that P_{0}Q_{0} ≤ PQ for all pairs of points (P,Q) with P on C_{1} and Q on C_{2}. (2000  10 Marks)
Ans. Sol. Given that C_{1} : x^{2} = y – 1 ; C_{2} : y^{2 }= x – 1
Let P ( x_{1} ,x_{1}^{2} + 1) on C_{1} and Q ( y_{2}^{2} + 1,y^{2}) on C_{2}.
Now the reflection of pt P in the line y = x can be obtained by interchanging the values of abscissa and ordiante.
Thus reflection of pt. P ( x_{1},x_{1}^{2 }+ 1) is P_{1} ( x_{1}^{2} + 1,x_{1}) and reflection of pt. Q ( y_{2}^{2 }+ 1,y^{2}) is Q_{1} ( y^{2} ,y_{2}^{2} + 1)
It can be seen clearly that P_{1} lies on C_{2 }and Q_{1} on C_{1}.
Now PP_{1} and QQ_{1} both are perpendicular to mirror line y = x.
Also M is mid pt. of PP_{1} (Q P_{1} is morror image of P in y = x)
In rt ΔPML,
PL > PM ⇒..(i)
Similarly,
...(ii)
Adding (i) and (ii) we get
⇒ PQ is more than the mean of PP_{1} and QQ_{1 }
⇒ PQ ≥ min (PP_{1}, QQ_{1}) Let min (PP_{1}, QQ_{1}) = PP_{1}
then
= 2(x_{1}^{2} + 1 – x_{1})^{2} = f (x1)
⇒ f '( x_{1}) = 4(x_{1}^{2 }+ 1 – x_{1})(2x_{1} – 1)
∴f ' (x_{1}) = 0 when x_{1} =
Also f ' (x_{1}) < 0 if x_{1}
⇒ f (x_{1}) is min when x_{1} =
Thus if at x_{1} = pt P is P_{0 }on C_{1}
Similarly Q_{0} on C_{2} will be image of P_{0} with respect to y = x
∴
Q.15. Let P be a point on the ellipse 0 < b < a. Let the line parallel to yaxis passing through P meet the circle x^{2} + y^{2} = a^{2} at the point Q such that P and Q are on the same side of xaxis. For two positive real numbers r and s, find the locus of the point R on PQ such that PR : RQ = r : s as P varies over the ellipse. (2001  4 Marks)
Ans.
Sol. Let the coordinates of P be (a cosθ , b sinθ ) then coordinates of Q are (a cosθ , a sinθ )
As R (h, k) divides PQ in the ratio r : s, then
⇒
⇒ ∵ cos^{2}θ + sin^{2}θ = 1
Hence locus of R is which is equation of an ellipse.
Q.16. Prove that, in an ellipse, the perpendicular from a focus upon any tangent and the line joining the centre of the ellipse to the point of contact meet on the corresponding directrix. (2002  5 Marks)
Ans. Sol.
Let the ellipse be and O be the centre.
Tangent at P (x_{1}, y_{1}) is = 0 whose
slope = Focus is S (ae, 0).
Equation of the line perpendicular to tangent at P is
....(1)
Equation of OP is ...(2)
(1) and (2) intersect
⇒ x (a^{2 }– b^{2}) = a^{3}e ⇒ x. a^{2}e^{2} = a^{3}e
⇒ x = a/e
Which is the corresponding directrix.
Q.17. Normals are drawn from the point P with slopes m_{1}, m_{2}, m_{3} to the parabola y^{2} = 4x. If locus of P with m_{1} m_{2 }= a is a part of the parabola itself then find a. (2003  4 Marks)
Ans. α = 2
Sol. Let P be the pt. (h, k). Then eqn of normal to parabola y^{2} = 4x from point (h, k), if m is the slope of normal, is y = mx – 2m – m^{3}
As it passes through (h, k), therefore mh – k – 2m – m^{3} = 0 or, m^{3} + (2 – h) + k = 0 ...(1)
Whcih is cubic in m, giving three values of m say m_{1}, m_{2 }and m_{3}.
Then m_{1}m_{2}m_{3} = – k (from eqn) but given that m_{1}m_{2} = a
∴We get m_{3}
But m^{3 }must satisfy eq^{n} (1)
⇒ k^{2} – 2α^{2} – hα^{2} – α3 = 0
∴Locus of P (h, k) is y^{2} = α^{2}x + (α^{3} – 2α^{2})
But ATQ, locus of P is a part of parabola y^{2} = 4x, therefore comparing the two, we get α^{2} = 4 and α^{3} – 2α^{2} = 0 ⇒ α = 2
Q.18. Tangent is drawn to parabola y^{2} – 2y – 4x + 5 = 0 at a point P which cuts the directrix at the point Q. A point R is such that it divides QP externally in the ratio 1/2 : 1. Find the locus of point R. (2004  4 Marks)
Ans. ( x  1)(y  1)^{2} + 4=0
Sol. The given eq^{n} of parabola is y^{2} – 2y – 4x + 5 = 0 ...(1)
⇒ (y – 1)^{2} = 4 (x – 1)
Any parametric point on this parabola is P (t^{2} + 1, 2t + 1)
Differentiating (1) w.r. to x, we get
∴Slope of tangent to (1) at point P (t^{2} + 1, 2t + 1) is
∴Eqn of tangent at P (t^{2} + 1, 2t + 1) is
y – (2t + 1)
⇒ yt – 2t^{2} – t = x – t^{2} = 1
⇒ x – yt + (t^{2} + t – 1) = 0 ...(2)
Now directrix of given parabola is (x – 1) = – 1 ⇒ x = 0
Tangent (2) meets directix at
Let pt. R be (h, k)
ATQ, R divides the line joing QP in the ratio :1 i.e., 1 : 2
externally.
⇒ h = – (1 + t^{2}) and
⇒ t^{2} = – 1 – h and
Eliminating t, we get
⇒ 4 = – (1 – k)^{2} (1 – h) ⇒ (h – 1) (k – 1)^{2 }+ 4 = 0
∴Locus of R (h, k) is (x – 1) (y – 1)^{2 }+ 4 = 0
Q.19. Tangents are drawn from any point on the hyperbola
to the circle x^{2} + y^{2} = 9. Find the locus of midpoint of the chord of contact. (2005  4 Marks)
Ans.
Sol. Any pt on the hyperbola is (3 secθ , 2 tanθ) Then, equation of chord of contact to the circle x^{2} + y^{2 }= 9,
with respect to the pt. (3 secθ, 2 tanθ) is (3 secθ) x + (2 tanθ) y = 9 ...(i)
If (h, k) be the mid point of chord of contact then equation of chord of contact will be hx + ky – 9 = h^{2}^{ }+ k^{2} – 9 (T = S_{1})
or, hx + ky = h^{2} + k^{2} ...(ii)
But equations (i) and (ii) represent the same st. line and hence should be identical, therefore, we get
sec^{2}θ – tan^{2}θ = 1
∴Locus of (h, k) is
Q.20. Find the equation of the common tangent in 1^{st} quadrant to the circle x^{2} + y^{2} = 16 and the ellipse
Also find the length of the intercept of the tangent between the coordinate axes. (2005  4 Marks)
Ans.
Sol. Let the common tangent to circle x^{2} + y^{2} = 16 and ellipse x^{2}/25 + y^{2} /4 = 1 be
..(i)
As it is tangent to circle x^{2} + y^{2} = 16, we should have
[Using : length of perpendicular from (0,0 ) to (1) = 4]
⇒ 25m^{2} + 4 = 16m^{2} + 16 ⇒ 9m^{2 }=12
[Leaving + ve sign to consider tangent in I quadrant] ∴Equation of common tangent is
This tangent meets the axes at and
∴Length of intercepted portion of tangent between axes
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132 docs70 tests
