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# Notes | EduRev

## JEE : Notes | EduRev

The document Notes | EduRev is a part of the JEE Course JEE Revision Notes.
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Illustration 1: Tangent is drawn to ellipse x2/27 + y2 = 1 at (3√3cos θ, sin θ) (where θ ∈ (0, π/2)). Then what is the value of θ for which the sum of intercepts on axis made by this tangent is minimum?
Solution: The given equation of parabola is x2/27 + y2 = 1
A tangent is drawn at the point (3√3cos θ, sin θ).
Hence, the equation of tangent at this point is given by
(x cos θ)/ 3√3 + (y sin θ)/ 1 = 1
Thus, the sum of intercepts = [3√3/ (cos θ) + 1/ (sin θ)] = f (θ) (say)
Hence, f’ (θ) = (3√3 sin3θ – cos3θ)/ (sin2θ – cos2θ)
Now, equating f’ (θ) = 0, we get
sin3θ = cos3θ/33/2
Hence, tan θ = 1/√3, i.e. θ = π/6
And so, at θ = π/6, f” (0) > 0
Hence, this implies that the tangent is minimum at π/6.

Illustration 2: Prove that, in an ellipse, the perpendicular from a focus upon any tangent and the line joining the center of the ellipse of the point of contact meet on the corresponding directrix. (2002)
Solution: Let the equation of the ellipse be x2/a+ y2/b2 = 1
Then any point on the ellipse is of the form P (a cos θ, b sin θ).
The equation of tangent at point P is given by
(x cos θ) /a + (y sin θ) /b = 1
The equation of line perpendicular to tangent is
(x sin θ) /b - (y cos θ) /a = μ
Now, since this passes through the focus (ae, 0), so
(ae sin θ)/b – 0 = μ
This gives μ = (ae sin θ)/b
Therefore the equation is (x sin θ) /b - (y cos θ) /a = (ae sin θ)/b    …. (1)
Equation of the line joining center and point of contact P (a cos θ, b sin θ) is
y = b/a (tan θ)x … (2)
Point of intersection Q of equations (1) and (2) has x coordinate a/e.
Hence, Q lies on the corresponding directrix x = a/e.

Illustration 3: Find the equation of the common tangent in 1st quadrant to the circle x2 + y2 = 16 and the ellipse x2/25 + y2/ 4 = 1. Also find the length of the intercept of the tangent between the coordinate axis. (2005)
Solution: Let the common tangent to x2 + y2 = 16 and x2/25 + y2/ 4 = 1 be y = mx + 4 √(1 + m2) And y = mx + √(25m2 + 4)
Since both these represent the same tangent, so
4 √(1+m2) = √(25m2 + 4)
So, 16 (1 + m2) = 25m2 + 4
Hence, 9m2 = 12
Which gives m = ± 2/√3
Since, tangent is in first quadrant, so m < 0
Hence, m = -2/√3
So, the equation of the common tangent is y = -2x/√3 + 4√7/3
Which meets the coordinate axis at A (2√7, 0) and (0, 4 √7/3).
Hence, AB = √ [(2√7 – 0)+ (0 – 4 √7/3)2]
= √ 28 + 11/3
= √196/3
= 14/√3
=14√3/3

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