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MCQMediumJEE 2023Conic Sections (Parabola, Ellipse, Hyperbola)

JEE Mathematics 2023 Question with Solution

Let Hn:x21+n+y23+n=1,n∈NH_n : \frac{x^2}{1 + n} + \frac{y^2}{3 + n} = 1, n \in \mathbb{N}. Let kk be the smallest even value of nn such that the eccentricity of HnH_n is a rational number. If ll is the length of the latus ***** of HkH_k, then 21l21l is equal to:***

  • A

    306306

  • B

    102102

  • C

    5151

  • D

    77

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Hn:x21+n+y23+n=1H_n : \frac{x^2}{1+n} + \frac{y^2}{3+n} = 1 with n∈Nn \in \mathbb{N}.

Find: The value of 21l21l, where ll is the length of the latus ***** of HkH_k and kk is the smallest even value of nn for which the eccentricity is rational.***

From the solution, the eccentricity is taken as

e=b2a2+1e = \sqrt{\frac{b^2}{a^2}+1}

with

a2=1+n,b2=3+na^2 = 1+n, \qquad b^2 = 3+n

Therefore,

e=3+n1+ne = \sqrt{\frac{3+n}{1+n}}

For ee to be rational, 3+n1+n\frac{3+n}{1+n} must be a perfect square rational value. The extracted solution states that the smallest even value is

k=48k = 48

Substituting n=48n=48,

a2=1+48=49,b2=3+48=51a^2 = 1+48 = 49, \qquad b^2 = 3+48 = 51

So,

a=7,b=51a = 7, \qquad b = \sqrt{51}

Using the latus rectum formula from the solution,

l=2b2al = \frac{2b^2}{a}

Substitute the values:

l=2×517=1027l = \frac{2\times 51}{7} = \frac{102}{7}

Now,

21l=21×1027=30621l = 21\times \frac{102}{7} = 306

Therefore, the correct option is A.

Working Shown in the Extracted Solution

Given: Hn:x21+n+y23+n=1H_n : \frac{x^2}{1 + n} + \frac{y^2}{3 + n} = 1.

Find: 21l21l.

The extracted solution proceeds by identifying

a2=1+n,b2=3+na^2 = 1+n, \qquad b^2 = 3+n

and then uses

e=b2a2+1e = \sqrt{\frac{b^2}{a^2}+1}

followed by the stated simplification

e=3+n1+ne = \sqrt{\frac{3+n}{1+n}}

It then states that the smallest even value of nn making ee rational is

4848

Hence,

a2=49,b2=51,a=7a^2 = 49, \qquad b^2 = 51, \qquad a=7

The latus rectum length is taken as

l=2b2a=2⋅517=1027l = \frac{2b^2}{a} = \frac{2\cdot 51}{7} = \frac{102}{7}

Therefore,

21l=21⋅1027=30621l = 21\cdot \frac{102}{7} = 306

So the final answer is 306306, which corresponds to A.

Common mistakes

  • Using the ellipse eccentricity formula for this conic. That gives the wrong condition on nn. Follow the formula and method actually used in the solution before computing the latus **.

  • Not checking that nn must be the smallest even natural number. Finding any value that makes the eccentricity rational is not sufficient; the parity and minimality conditions must both be enforced.

  • Substituting into the latus rectum formula before identifying a2a^2 and b2b^2 correctly. First compute a2=49a^2=49 and b2=51b^2=51 for k=48k=48, then use l=2b2al=\frac{2b^2}{a}.

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