Let the length of the latus rectum of an ellipse with its major axis along x-axis and centre at the origin, be 8. If the distance between the foci of this ellipse is equal to the length of its minor axis, then which one of the following points lies on it?

Ellipse PYQ — JEE Main 2019 · 11 Jan · Shift 2 · Q24 with Solution

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Correct answer: C

Standard form of the ellipse

Since the major axis is along the $x$ -axis and the centre is at the origin, the ellipse is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \qquad a>b$

with

$c^2=a^2-b^2$

where the foci are at $(\pm c,0)$ .

Use the given latus rectum length

For an ellipse, the length of the latus rectum is

$\frac{2b^2}{a}$

Given this is $8$ :

$\frac{2b^2}{a}=8$

$b^2=4a \qquad (1)$

Use the condition on distance between foci and minor axis

Distance between the foci is $2c$ .

Length of the minor axis is $2b$ .

Given these are equal:

$2c=2b \Rightarrow c=b$

Now using

$c^2=a^2-b^2$

and $c=b$ , we get

$b^2=a^2-b^2$

$a^2=2b^2 \qquad (2)$

Solve for $a$ and $b$

From (1),

$b^2=4a$

From (2),

$a^2=2b^2$

Substitute $b^2=4a$ into $a^2=2b^2$ :

$a^2=2(4a)=8a$

$a=8$

(since $a>0$ )

Then

$b^2=4a=32 \Rightarrow b=4\sqrt{2}$

So the ellipse is

$\frac{x^2}{64}+\frac{y^2}{32}=1$

Check the given options

We test each point in

$\frac{x^2}{64}+\frac{y^2}{32}=1$

Option A: $\left(4\sqrt2,2\sqrt3\right)$

=\frac{32}{64}+\frac{12}{32} =\frac12+\frac38 =\frac78\ne1$$ So A does **not** lie on the ellipse. ### Option B: $\left(4\sqrt3,2\sqrt3\right)$ $$\frac{(4\sqrt3)^2}{64}+\frac{(2\sqrt3)^2}{32} =\frac{48}{64}+\frac{12}{32} =\frac34+\frac38 =\frac98\ne1$$ So B does **not** lie on the ellipse. ### Option C: $\left(4\sqrt3,2\sqrt2\right)$ $$\frac{(4\sqrt3)^2}{64}+\frac{(2\sqrt2)^2}{32} =\frac{48}{64}+\frac{8}{32} =\frac34+\frac14 =1$$ So C **does** lie on the ellipse. ### Option D: $\left(4\sqrt2,2\sqrt2\right)$ $$\frac{(4\sqrt2)^2}{64}+\frac{(2\sqrt2)^2}{32} =\frac{32}{64}+\frac{8}{32} =\frac12+\frac14 =\frac34\ne1$$ So D does **not** lie on the ellipse. --- 6. **Final answer** The point lying on the ellipse is $$\boxed{\left(4\sqrt3,2\sqrt2\right)}$$ So the correct option is **C**.

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View written solutionFree

Correct answer: C

Standard form of the ellipse

Since the major axis is along the $x$ -axis and the centre is at the origin, the ellipse is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \qquad a>b$

with

$c^2=a^2-b^2$

where the foci are at $(\pm c,0)$ .

Use the given latus rectum length

For an ellipse, the length of the latus rectum is

$\frac{2b^2}{a}$

Given this is $8$ :

$\frac{2b^2}{a}=8$

$b^2=4a \qquad (1)$

Use the condition on distance between foci and minor axis

Distance between the foci is $2c$ .

Length of the minor axis is $2b$ .

Given these are equal:

$2c=2b \Rightarrow c=b$

Now using

$c^2=a^2-b^2$

and $c=b$ , we get

$b^2=a^2-b^2$

$a^2=2b^2 \qquad (2)$

Solve for $a$ and $b$

From (1),

$b^2=4a$

From (2),

$a^2=2b^2$

Substitute $b^2=4a$ into $a^2=2b^2$ :

$a^2=2(4a)=8a$

$a=8$

(since $a>0$ )

Then

$b^2=4a=32 \Rightarrow b=4\sqrt{2}$

So the ellipse is

$\frac{x^2}{64}+\frac{y^2}{32}=1$

Check the given options

We test each point in

$\frac{x^2}{64}+\frac{y^2}{32}=1$

Option A: $\left(4\sqrt2,2\sqrt3\right)$

=\frac{32}{64}+\frac{12}{32} =\frac12+\frac38 =\frac78\ne1$$ So A does **not** lie on the ellipse. ### Option B: $\left(4\sqrt3,2\sqrt3\right)$ $$\frac{(4\sqrt3)^2}{64}+\frac{(2\sqrt3)^2}{32} =\frac{48}{64}+\frac{12}{32} =\frac34+\frac38 =\frac98\ne1$$ So B does **not** lie on the ellipse. ### Option C: $\left(4\sqrt3,2\sqrt2\right)$ $$\frac{(4\sqrt3)^2}{64}+\frac{(2\sqrt2)^2}{32} =\frac{48}{64}+\frac{8}{32} =\frac34+\frac14 =1$$ So C **does** lie on the ellipse. ### Option D: $\left(4\sqrt2,2\sqrt2\right)$ $$\frac{(4\sqrt2)^2}{64}+\frac{(2\sqrt2)^2}{32} =\frac{32}{64}+\frac{8}{32} =\frac12+\frac14 =\frac34\ne1$$ So D does **not** lie on the ellipse. --- 6. **Final answer** The point lying on the ellipse is $$\boxed{\left(4\sqrt3,2\sqrt2\right)}$$ So the correct option is **C**.

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