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 12 d equals \underline\hspace2cm .

Parabola PYQ — JEE Main 2024 · 6 Apr · Shift 1 · Q52 with Solution

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Correct answer: 75

Interpret the geometric condition

Let $P(x,y)$ be any point on the conic $C$ .

Slope of tangent at $P$ is $\dfrac{dy}{dx}$ .
Slope of the line joining $P(x,y)$ and $(3,-5)$ is $\frac{y-(-5)}{x-3}=\frac{y+5}{x-3}.$

Given: $\frac{dy}{dx}=\frac12\cdot \frac{y+5}{x-3}.$

So the curve satisfies the differential equation $\frac{dy}{dx}=\frac{y+5}{2(x-3)}.$

Solve the differential equation

Separate variables: $\frac{dy}{y+5}=\frac{dx}{2(x-3)}.$

Integrate: $\int \frac{dy}{y+5}=\int \frac{dx}{2(x-3)}$ $\ln|y+5|=\frac12\ln|x-3|+C.$

Exponentiating, $y+5=k\sqrt{x-3}$ for some constant $k$ .

Squaring, $(y+5)^2=k^2(x-3).$

This is a parabola with vertex $(3,-5)$ opening rightward.

Use the point $(4,-2)$ to determine the parabola

Since $(4,-2)$ lies on $C$ , $(-2+5)^2=k^2(4-3)$ $3^2=k^2$ $k^2=9.$

Hence the parabola is $(y+5)^2=9(x-3).$

Compare with standard form $(y-k)^2=4a(x-h).$

So here, $h=3,\quad k=-5,\quad 4a=9 \implies a=\frac94.$

Thus the focus is $(h+a,k)=\left(3+\frac94,-5\right)=\left(\frac{21}{4},-5\right).$

Compute focal distance of the point $(7,1)$ on the parabola

Check that $(7,1)$ lies on the parabola:

\qquad 9(7-3)=36.$$ So yes. Now the focal distance $d$ is the distance from $(7,1)$ to the focus $\left(\frac{21}{4},-5\right)$: $$d=\sqrt{\left(7-\frac{21}{4}\right)^2+(1+5)^2}.$$ Compute: $$7-\frac{21}{4}=\frac{28-21}{4}=\frac74.$$ Hence $$d=\sqrt{\left(\frac74\right)^2+6^2} =\sqrt{\frac{49}{16}+36} =\sqrt{\frac{49+576}{16}} =\sqrt{\frac{625}{16}} =\frac{25}{4}.$$ Therefore, $$12d=12\cdot \frac{25}{4}=75.$$ 5. **Final answer** $$\boxed{75}$$ The derived answer matches the stored correct answer.

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View written solutionFree

Correct answer: 75

Interpret the geometric condition

Let $P(x,y)$ be any point on the conic $C$ .

Slope of tangent at $P$ is $\dfrac{dy}{dx}$ .
Slope of the line joining $P(x,y)$ and $(3,-5)$ is $\frac{y-(-5)}{x-3}=\frac{y+5}{x-3}.$

Given: $\frac{dy}{dx}=\frac12\cdot \frac{y+5}{x-3}.$

So the curve satisfies the differential equation $\frac{dy}{dx}=\frac{y+5}{2(x-3)}.$

Solve the differential equation

Separate variables: $\frac{dy}{y+5}=\frac{dx}{2(x-3)}.$

Integrate: $\int \frac{dy}{y+5}=\int \frac{dx}{2(x-3)}$ $\ln|y+5|=\frac12\ln|x-3|+C.$

Exponentiating, $y+5=k\sqrt{x-3}$ for some constant $k$ .

Squaring, $(y+5)^2=k^2(x-3).$

This is a parabola with vertex $(3,-5)$ opening rightward.

Use the point $(4,-2)$ to determine the parabola

Since $(4,-2)$ lies on $C$ , $(-2+5)^2=k^2(4-3)$ $3^2=k^2$ $k^2=9.$

Hence the parabola is $(y+5)^2=9(x-3).$

Compare with standard form $(y-k)^2=4a(x-h).$

So here, $h=3,\quad k=-5,\quad 4a=9 \implies a=\frac94.$

Thus the focus is $(h+a,k)=\left(3+\frac94,-5\right)=\left(\frac{21}{4},-5\right).$

Compute focal distance of the point $(7,1)$ on the parabola

Check that $(7,1)$ lies on the parabola:

\qquad 9(7-3)=36.$$ So yes. Now the focal distance $d$ is the distance from $(7,1)$ to the focus $\left(\frac{21}{4},-5\right)$: $$d=\sqrt{\left(7-\frac{21}{4}\right)^2+(1+5)^2}.$$ Compute: $$7-\frac{21}{4}=\frac{28-21}{4}=\frac74.$$ Hence $$d=\sqrt{\left(\frac74\right)^2+6^2} =\sqrt{\frac{49}{16}+36} =\sqrt{\frac{49+576}{16}} =\sqrt{\frac{625}{16}} =\frac{25}{4}.$$ Therefore, $$12d=12\cdot \frac{25}{4}=75.$$ 5. **Final answer** $$\boxed{75}$$ The derived answer matches the stored correct answer.

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