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

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Solution

Q.
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 ________ . [2023]

Ans.
(292)

Equation of tangent at P(1, 3) to the curve $x^{2} + 2 x - 4 y + 9 = 0$ is

$x \cdot 1 + 2 (\frac{1 + x}{2}) - 4 (\frac{3 + y}{2}) + 9 = 0$

$\Rightarrow x + 1 + x - 6 - 2 y + 9 = 0$

$\Rightarrow 2 x - 2 y + 4 = 0 \Rightarrow y - x = 2$

So, the point $A$ is $(0, 2)$ .

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

Point $(1, 3)$ lies on this line.

$∴$    $1 - 9 = d \Rightarrow d = - 8$

So, equation of line is $x - 3 y = - 8$ .

Now, $x - 3 y = - 8$ meets the parabola $y^{2} = 4 x$ at $B$ .

So, $\frac{x + 8}{3} = y \Rightarrow {(\frac{x + 8}{3})}^{2} = 4 x$ $\Rightarrow x^{2} + 16 x + 64 = 36 x$

$\Rightarrow x^{2} - 16 x - 4 x + 64 = 0 \Rightarrow x = 4, 16$    $∴$ $y = 4, 8$

So, the possible coordinates of $B$ are $(4, 4)$ or $(16, 8)$ .

But $(4, 4)$ doesn't lie on the line $2 x - 3 y = 8$ .

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

$∴$    $A B^{2} = {(16 - 0)}^{2} + {(8 - 2)}^{2} = 256 + 36 = 292$

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