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  • If the tangent at  to the curve <img alt="4 x^2=y-8" src="https://entrancecorner.oncodecogs.com/g

If the tangent at (1,7) to the curve 4 x^2=y-8 touches the circle x^2+y^2+12 x+12 y+c=0 

then the value of c is 

 

Option: 1

-98


Option: 2

98


Option: 3

72


Option: 4

-72


Answers (1)

best_answer

The equation of the tangent to the curve 4 x^2=y-8 at the point (1,7) Differentiating4 x^2=y-8 with respect to x, 8x=dx/dy At the point (1,7) ,8(1)=8 Therefore, the equation of the tangent at (1,7) is,

\begin{aligned} & y-7=8(x-1) \\ & y=8 x-1 \end{aligned}

the circle,

\begin{gathered} x^2+y^2+12 x+12 y+c=0 \\ (x+6)^2+(y+6)^2-36-36+c=0 \\ (x+6)^2+(y+6)^2=72-c \end{gathered}

This is the equation of a circle with center (-6,-6),(1,7)is,

\begin{aligned} & \left.\quad \sqrt{(}(1-(-6))^2+(7-(-6))^2\right)=\operatorname{sqrt}(170) \\ & \sqrt{(}(72-c)=\sqrt{(170)} \\ & 72-c=170 \end{aligned}

Therefore, the value of c is-98

Posted by

rishi.raj

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