Let the tangent to the curve x^2+2 x-4 y+9=0 at the point P (1,3)… - Vidyadip

MCQ

Let the tangent to the curve $x^2+2 x-4 y+9=0$ at the point $P (1,3)$ on it meet the $y$-axis at $A$. Let the line passing through $P$ and parallel to the line $x -$ $3 y=6$ meet the parabola $y^2=4 x$ at $B$. If $B$ lies on the line $2 x-3 y=8$. then $(A B)^2$ is equal to $............$.

A
$291$
B
$290$
C
$293$
✓
$292$

✓

Answer

Correct option: D.

$292$

d
Equation of tangent at $P(1,3)$ to the curve

$x ^2+2 x -4 y +9=0 \text { is } y - x =2$

Then the point $A$ is $(0,2)$

Equation of line passing through $P$ and parallel to the line $x -3 y =6$.

The possible coordinate of $B$ are $(4,4)$ or $(16,8)$

But $(4,4)$ does not satisfy $2 x-3 y=8$

Thus the point $B$ is $(16,8)$

Then $( AB )^2=292$

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