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Parabola question

2024 · Shift 2 · Q29
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Parabola question

2024 · Shift 2 · Q29

JEE AdvancedMathematicsParabolaNumerical+4 / −1
A normal with slope 16\frac{1}{\sqrt{6}}6​1​ is drawn from the point (0,−α)(0,-\alpha)(0,−α) to the parabola x2=−4ayx^2=-4 a yx2=−4ay, where a>0a\gt 0a>0. Let LLL be the line passing through (0,−α)(0,-\alpha)(0,−α) and parallel to the directrix of the parabola. Suppose that LLL intersects the parabola at two points AAA and BBB. Let rrr denote the length of the latus rectum and sss denote the square of the length of the line segment ABA BAB. If r:s=1:16r: s=1: 16r:s=1:16, then the value of 24a24 a24a is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 12

  1. Given parabola and its standard properties

The parabola is x2=−4ay,a>0.x^2=-4ay, \qquad a>0.x2=−4ay,a>0.

This is of the form x2=−4ayx^2=-4ayx2=−4ay, so:

  • vertex: (0,0)(0,0)(0,0)
  • focus: (0,−a)(0,-a)(0,−a)
  • directrix: y=ay=ay=a
  • latus rectum length: r=4a.r=4a.r=4a.
  1. Equation of a normal to the parabola

For the parabola x2=−4ayx^2=-4ayx2=−4ay, a parametric point is P(2at,−at2).P(2at,-at^2).P(2at,−at2).

Differentiate implicitly:

\Rightarrow \frac{dy}{dx}=-\frac{x}{2a}.$$ At $P(2at,-at^2)$, slope of tangent is $$m_t=-t,$$ so slope of normal is $$m_n=\frac{1}{t}.$$ We are given that the normal has slope $$\frac{1}{\sqrt{6}},$$ therefore $$\frac{1}{t}=\frac{1}{\sqrt{6}} \Rightarrow t=\sqrt{6}.$$ --- 3. **Equation of the normal at parameter $t$** Normal at $P(2at,-at^2)$ is $$y+at^2=\frac{1}{t}(x-2at).$$ Substitute $t=\sqrt{6}$: $$y+6a=\frac{1}{\sqrt{6}}(x-2a\sqrt{6}).$$ Simplify: $$y+6a=\frac{x}{\sqrt{6}}-2a$$ $$\Rightarrow y=\frac{x}{\sqrt{6}}-8a.$$ This normal passes through $(0,-\alpha)$. Hence at $x=0$, $$-\alpha=-8a \Rightarrow \alpha=8a.$$ So the point is $$(0,-8a).$$ --- 4. **Line $L$ through $(0,-\alpha)$ parallel to the directrix** Since directrix is $y=a$, a line parallel to it is horizontal. Thus line $L$ is $$y=-\alpha=-8a.$$ --- 5. **Intersection of $L$ with the parabola** Substitute $y=-8a$ into $$x^2=-4ay$$ $$x^2=-4a(-8a)=32a^2.$$ So the intersection points are $$A(4\sqrt{2}a,-8a), \qquad B(-4\sqrt{2}a,-8a).$$ Therefore, $$AB=4\sqrt{2}a-(-4\sqrt{2}a)=8\sqrt{2}a.$$ Hence, $$s=(AB)^2=(8\sqrt{2}a)^2=128a^2.$$ --- 6. **Use the ratio $r:s=1:16$** We know $$r=4a, \qquad s=128a^2.$$ Given $$r:s=1:16,$$ so $$\frac{r}{s}=\frac{1}{16}.$$ Thus, $$\frac{4a}{128a^2}=\frac{1}{16}.$$ Simplify: $$\frac{1}{32a}=\frac{1}{16}$$ $$\Rightarrow 32a=16$$ $$\Rightarrow a=\frac{1}{2}.$$ Therefore, $$24a=24\cdot \frac{1}{2}=12.$$ --- 7. **Final answer** $$\boxed{12}$$ The derived answer matches the stored correct answer.
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