Find slope of normal to the curve y=5x^{2}10x + 7 at x=1
y = 5x^{2}  10x + 7
dy/dx = 10x  10
(At x = 1) 10(1)  10
m1 = 0
As we know that slope, m1m2 = 1
=> 0(m2) = 1
m2 = 1/0 (which is not defined)
The equation of the tangent line to the curve y = which is parallel to the line 4x 2y + 3 = 0 is
Find the equation of tangent to which has slope 2.
y = 1/(x3)
dy/dx = d/dx(x3)^{1}
dy/dx = (1) (x3)^{(11)} . d(x3)/dx
dy/dx = (x3)^{(2)}
dy/dx =  1/(x3)^{2}
Given, slope = 2, dy/dx = 2
 1/(x3)^{2} = 2
⇒ 1 = 2(x3)^{2}
2(x3)^{2} = 1
(x3)^{2} = ½
We know that square of any number is always positive So, (x3)^{2} > 0
(x3)^{2} = ½ is not possible
No tangent to the curve has slope 2.
If y = – sin2x. Find the values of x at which the tangents drawn to the graph of this function is parallel to the x axis.
The normal at any point q to the curve x = a (cos q + q sin q), y = a (sin q – q cos q) is at distance from the origin that is equal to… .
Clearly, dx/dy =tan q = slope of normal = −cotq
Equation of normal at θ' is y−a(sinq − qcosq) = −cotq(x−a(cosq + qsinq)
= ysinq − asin^{2}q + aqcosqsinq
= −xcosq + acos^{2}q + aqsinqcosq
= xcosq + ysinq = a
The equation of the normal to the curve x^{2} = 4y which passes through the point (1, 2) is.
h^{2 }= 4k
slope of normal=−1/(dy/dx) = −2h
equation of normal(y − k)= −2h(x−h)
k = 2 + 2/h(1 − h)
(h^{2}) / 4 = 2 + 2/h (1 − h)
h = 2, k = 1
equation of line (y  1)= 1(x  2)
x + y = 3
The curve y = ax^{3} + bx^{2} + cx + 5 touches the xaxis at P(2, 0) and cuts the yaxis at the point Q where its gradient is 3. the equation of the curve is… .
y=ax^{3}+bx^{2}+cx+5
(−z,0) lies on curve y=ax^{3}+ bx^{2} + cx + 5d
⇒ 0= −8a + 4b − 2c + 5.....(1)
Also at (−2,0) curve touches xaxis ie dy/dx at (−2,0) is 0
∴ dy/dx=3ax^{2}+2bx+c
dy/dx (−2,0)=0=12a = −4b + c....(2)
Since curve crosses yaxis at Q, xcoordinate of point Q must be zero
⇒ y = 0 + 0 + 0 + 5
⇒y = 5
∴ coordinate of Q = (0,5)
According to Question dy/dx =3 at Q(0,5)
⇒ 3= 0 + 0 + c
⇒ c = 3
putting the value of c in eqn(1) & eq n(2)and gurthare solving we get
a=−1/2 & b=3/4
∴Eqn of cover is y = −(x^{3}) / 2−3/4x^{2 }+ 3x + 5
If a, b are real numbers such that x^{3}ax^{2} + bx – 6 = 0 has its roots real and positive then minimum value of b is
If p,q and r are the roots,
pq + qr + pr = b and pqr = 6
Since A.M.≥G.M . applying it to 3 numbers pq,qr and pr
(pq + qr + pr)/3 ≥ (p^{2}q^{2}r^{2})^{1/3} i.e.
(pq + qr + pr)/3 ≥ (36)^{1/3}
pq + qr + pr ≥ 3 (36)^{1/3}
So minimum value of b is (36)^{1/3}
The equation of tangent to the curve y = x^{3} + 2x + 6 which is perpendicular to the line x + 14y + 4 = 0 is :
y = x^{3} + 2x + 6
Slope of tangent = m1 = dy/dx = 3x^{3} + 2
Slope of perpendicular line = m2 = −1/14
m1 . m2 = 1
m1 = 14
3x^{2} + 2 = 14
x = ±2
Therefore the curve has tangents at x = 2 and x = 2
and these points also lie on the given curve
Equation of tangent  y = 14x + c
Coordinates of points of tangency
At x = 2 , y = 23 + 2(2) + 6 = 18
At x = 2 , y = (2)3 + 2(2) + 6 = 6
18 = 14(2) + c
c = 10
6 = 14(2) + c
c = 22
Equation of tangent  y = 14x  10 and y = 14x + 22
Hence, 14x  y +22 = 0
The equations of the tangents drawn to the curve y^{2} – 2x^{3} – 4y + 8 = 0 from the point (1, 2) is… .
Let P(α, β) be any point on the curve
Now, the equation of the tangent at P is
Hence, the point of contacts are
(2, 2 + 2√3) and (2, 2 – 2√3)
Slope of the tangents are 2√3 , –2√3
Hence, the equations of tangents are
y – (2 + 2√3) = 2√3 (x – 2)
and
y –(2 – 2√3) = – 2√3 (x – 2)
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