JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Hard

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Analyzing +4 | -1

If the tangent to the conic,

y-6=x^2

(2,10)

touches the circle,

x^2+y^2+8 x-2 y=k

(for some fixed

k

) at a point

(\alpha, \beta)

; then

(\alpha, \beta)

is :

View answer and explanation

Correct option: D

\left(-\frac{8}{17}, \frac{2}{17}\right)

Option D (

\left(-\frac{8}{17}, \frac{2}{17}\right)

) is correct. The tangent to the conic touches the circle at a point that can be found by equating the slopes and solving the system of equations.

More Questions from TANGENTS & NORMALS

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a

y=x^3-a x

x^2 y=1-y

x y=1-y

C

y^3-3 x y+2=0

H

\mathrm{C}

x

V

C

y

y-e^{x y}+x=0

y = f\left( x \right)

x

x = a\left( {\theta - \sin \theta } \right),\;y = a\left( {1 + \cos \theta } \right)

\theta = \left( {2k + 1} \right)\pi ,\;k \in Z