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  • The tangent at  to the curves  <img alt="x^2=y-6 x" src="https://entrancecorner.oncodec

The tangent at (1,7) to the curves  x^2=y-6 x, touches the circle  x^2+y^2+16 x+12 y+c=0 at 

Option: 1

\begin{aligned} & (6,7) \\ \end{aligned}


Option: 2

(-6,7) \\


Option: 3

(6,-7) \\


Option: 4

(-6,-7)


Answers (1)

best_answer

The tangent at (1,7) to the parabola x^2=y-6 x \: \: {\text {is }}

\begin{aligned} & x(1)=\frac{1}{2}(y+7)-6 \ldots \ldots \ldots \ldots . .\left[\text { replacing } x^2 \rightarrow x x_1\right]\left[2 y \rightarrow y+y_1\right] \\ \Rightarrow & 2 x=y+7-12 \\ \Rightarrow & y=2 x+5 \ldots \ldots \ldots \ldots \ldots \ldots(i) \end{aligned}

which is also tangent to the circle x^2+y^2+16 x+12 y+c=0 \text { i.e. }

x^2+(2 x+5)^2+16 x+12(2 x+5)+C=0

must have equal roots i.e. \alpha=\beta

\begin{aligned} & \Rightarrow 5 x^2+60 x+85+c=0 \\ \Rightarrow & \alpha+\beta=\frac{-60}{5} \\ \Rightarrow & \alpha=-6 \\ \therefore & x=-6 \text { and } \therefore y=2 x+5=-7 \end{aligned}

Point of contact is (-6,-7)

Posted by

rishi.raj

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