Let P be a parabola with vertex (2,3) and directrix 2 x+y=6 . Let an ellipse E: \fracx^2a^2+\fracy^2b^2=1, a>b , of eccentricity \frac1\sqrt2 pass through the focus of the parabola P . Then, the square of the length of the latus rectum of E , is

Ellipse PYQ — JEE Main 2024 · 31 Jan · Shift 2 · Q36 with Solution

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Correct answer: B

Parabola data

The parabola has vertex $V=(2,3)$ and directrix $2x+y=6.$

For a parabola, the axis is perpendicular to the directrix. Also, the vertex is midway between the focus and the directrix.

Distance from vertex to directrix

Write the directrix as $2x+y-6=0.$

Distance of vertex $(2,3)$ from this line is $d=\frac{|2\cdot 2+1\cdot 3-6|}{\sqrt{2^2+1^2}}=\frac{|4+3-6|}{\sqrt{5}}=\frac{1}{\sqrt{5}}.$

Hence the focal length of the parabola is $p=\frac{1}{\sqrt{5}}.$

Find the focus of the parabola

A normal vector to the directrix $2x+y-6=0$ is $(2,1)$ , whose unit vector is $\left(\frac{2}{\sqrt5},\frac{1}{\sqrt5}\right).$

Check which side of the directrix contains the vertex: $2(2)+3-6=1>0,$ so from the foot on the directrix to the vertex, the direction is along $\left(\frac{2}{\sqrt5},\frac{1}{\sqrt5}\right)$ . Therefore the focus lies the same distance beyond the vertex in this direction: $F=V+p\left(\frac{2}{\sqrt5},\frac{1}{\sqrt5}\right).$

Since $p=\frac1{\sqrt5}$ , $F=(2,3)+\left(\frac{2}{5},\frac{1}{5}\right)=\left(\frac{12}{5},\frac{16}{5}\right).$

Ellipse data

Given $E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad a>b,$ with eccentricity $e=\frac{1}{\sqrt2}.$

For an ellipse, $e^2=1-\frac{b^2}{a^2}.$ So

\quad\Rightarrow\quad \frac{b^2}{a^2}=\frac12 \quad\Rightarrow\quad b^2=\frac{a^2}{2}.$$ --- 5. **Ellipse passes through the focus of parabola** Since $\left(\frac{12}{5},\frac{16}{5}\right)$ lies on the ellipse, $$\frac{(12/5)^2}{a^2}+\frac{(16/5)^2}{b^2}=1.$$ Substitute $b^2=\frac{a^2}{2}$: $$\frac{144/25}{a^2}+\frac{256/25}{a^2/2}=1.$$ Now, $$\frac{256/25}{a^2/2}=\frac{512}{25a^2}.$$ Thus $$\frac{144}{25a^2}+\frac{512}{25a^2}=1 \quad\Rightarrow\quad \frac{656}{25a^2}=1.$$ Hence $$a^2=\frac{656}{25}.$$ Then $$b^2=\frac{a^2}{2}=\frac{328}{25}.$$ --- 6. **Length of latus rectum of the ellipse** For ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, length of latus rectum is $$L=\frac{2b^2}{a}.$$ So its square is $$L^2=\left(\frac{2b^2}{a}\right)^2=\frac{4b^4}{a^2}.$$ Using $b^2=\frac{a^2}{2}$, $$b^4=\frac{a^4}{4},$$ therefore $$L^2=\frac{4\cdot (a^4/4)}{a^2}=a^2.$$ But we found $$a^2=\frac{656}{25}.$$ Hence, $$\boxed{L^2=\frac{656}{25}}.$$ So the correct option is **B**.

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View written solutionFree

Correct answer: B

Parabola data

The parabola has vertex $V=(2,3)$ and directrix $2x+y=6.$

For a parabola, the axis is perpendicular to the directrix. Also, the vertex is midway between the focus and the directrix.

Distance from vertex to directrix

Write the directrix as $2x+y-6=0.$

Distance of vertex $(2,3)$ from this line is $d=\frac{|2\cdot 2+1\cdot 3-6|}{\sqrt{2^2+1^2}}=\frac{|4+3-6|}{\sqrt{5}}=\frac{1}{\sqrt{5}}.$

Hence the focal length of the parabola is $p=\frac{1}{\sqrt{5}}.$

Find the focus of the parabola

A normal vector to the directrix $2x+y-6=0$ is $(2,1)$ , whose unit vector is $\left(\frac{2}{\sqrt5},\frac{1}{\sqrt5}\right).$

Since $p=\frac1{\sqrt5}$ , $F=(2,3)+\left(\frac{2}{5},\frac{1}{5}\right)=\left(\frac{12}{5},\frac{16}{5}\right).$

Ellipse data

Given $E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad a>b,$ with eccentricity $e=\frac{1}{\sqrt2}.$

For an ellipse, $e^2=1-\frac{b^2}{a^2}.$ So

\quad\Rightarrow\quad \frac{b^2}{a^2}=\frac12 \quad\Rightarrow\quad b^2=\frac{a^2}{2}.$$ --- 5. **Ellipse passes through the focus of parabola** Since $\left(\frac{12}{5},\frac{16}{5}\right)$ lies on the ellipse, $$\frac{(12/5)^2}{a^2}+\frac{(16/5)^2}{b^2}=1.$$ Substitute $b^2=\frac{a^2}{2}$: $$\frac{144/25}{a^2}+\frac{256/25}{a^2/2}=1.$$ Now, $$\frac{256/25}{a^2/2}=\frac{512}{25a^2}.$$ Thus $$\frac{144}{25a^2}+\frac{512}{25a^2}=1 \quad\Rightarrow\quad \frac{656}{25a^2}=1.$$ Hence $$a^2=\frac{656}{25}.$$ Then $$b^2=\frac{a^2}{2}=\frac{328}{25}.$$ --- 6. **Length of latus rectum of the ellipse** For ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, length of latus rectum is $$L=\frac{2b^2}{a}.$$ So its square is $$L^2=\left(\frac{2b^2}{a}\right)^2=\frac{4b^4}{a^2}.$$ Using $b^2=\frac{a^2}{2}$, $$b^4=\frac{a^4}{4},$$ therefore $$L^2=\frac{4\cdot (a^4/4)}{a^2}=a^2.$$ But we found $$a^2=\frac{656}{25}.$$ Hence, $$\boxed{L^2=\frac{656}{25}}.$$ So the correct option is **B**.

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