Let a conic C pass through the point (4, –2) and P(x, y), be any point on C. Let the slope of the line touching the conic C only at a single point P be half the slope of the line joining the points P and (3, –5). If the focal distance of the point (7, 1)

Question

Let a conic C pass through the point (4, –2) and P(x, y), $x \geq 3$ , be any point on C. Let the slope of the line touching the conic C only at a single point P be half the slope of the line joining the points P and (3, –5). If the focal distance of the point (7, 1) on C is d, then 12d equals __________. [2024]

Smart Achievers · Accepted Answer

(75)

Slope of C at P $= \frac{1}{2} (\frac{y + 5}{x - 3})$

$\Rightarrow \frac{d y}{d x} = \frac{1}{2} (\frac{y + 5}{x - 3}) \Rightarrow \frac{2 d y}{y + 5} = \frac{1}{x - 3} d x$

On intergrating, we get

2 ln (y + 5) = ln (x – 3) + C

Since, C passes through (4, –2)

$\Rightarrow 2 \ln 3 = C$

$\Rightarrow 2 \ln (y + 5) = \ln (x - 3) + 2 \ln 3$

$\Rightarrow 2 \ln (\frac{y + 5}{3}) = \ln (x - 3) \Rightarrow {(\frac{y + 5}{3})}^{2} = x - 3$

$\Rightarrow {(y + 5)}^{2} = 9 (x - 3)$ , which represent a parabola so, 4a = 9

$\Rightarrow a = \frac{9}{4}$

Focus $= (3 + \frac{9}{4}, - 5) = (\frac{21}{4}, - 5)$

$∴ d = \sqrt{{(\frac{21}{4} - 7)}^{2} + {(- 5 - 1)}^{2}}$

$= \sqrt{{(- \frac{7}{4})}^{2} + {(- 6)}^{2}} = \frac{25}{4}$

$\Rightarrow 12 d = \frac{25}{4} \times 12 = 75$ .

Solution

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