$A(-2,9)$ and $B(1,6)$ are two points on the curve $y=x^2+5$. The coordinates of the point $\mathrm{C}$ on…
$A(-2,9)$ and $B(1,6)$ are two points on the curve $y=x^2+5$. The coordinates of the point $\mathrm{C}$ on the curve such that the tangent drawn at $\mathrm{A}$ is parallel to the chord $\mathrm{BC}$, is
$(-5,30)$
$(0,5)$
$(-9,86)$
$(6,41)$
Solution
$
\begin{aligned}
& \text {} \because y=x^2+5 \\
& \Rightarrow y^{\prime}=2 x
\end{aligned}
$
Equation of tangent line at point $A(-2,9)$ :
$
\begin{aligned}
& (y-9)=(2 x-2)(x+2) \\
& \Rightarrow y-9=-4 x-8 \Rightarrow y=-4 x+1
\end{aligned}
$
let coordinate of $\mathrm{C}$ is $\left(\mathrm{x}^{\prime}, \mathrm{y}^{\prime}\right)$
Then, $\frac{6-y^{\prime}}{1-x^{\prime}}=-4 \Rightarrow 6-y^{\prime}=-4+4 x^{\prime}$
$
\Rightarrow y^{\prime}+4 x^{\prime}=10
$
$\because \quad \mathrm{C}\left(\mathrm{x}^{\prime}, \mathrm{y}^{\prime}\right)$ lies on equation (i), we get
$
\mathrm{y}^{\prime}=\mathrm{x}^{\prime 2}+5
$
Using $\mathrm{eq}^{\mathrm{n}}$ (iv) in equation (iii), we get
$
\begin{aligned}
& \mathrm{x}^{\prime 2}+5+4 \mathrm{x}^{\prime}=10 \Rightarrow \mathrm{x}^{\prime 2}+4 \mathrm{x}^{\prime}-5=0 \Rightarrow \mathrm{x}^{\prime}=-5,1 \\
& \text { when } \mathrm{x}^{\prime}=-5 \Rightarrow \mathrm{y}^{\prime}=25+5=30 \\
& \text { Then } \mathrm{C} \equiv(-5,30)
\end{aligned}
$