Question

Let the centre of an ellipse is at $(0,0)$
Assertion: If major axis is on the y-axis and ellipse passes through the points $(3,2)$ and $(1,6)$, then the equation of ellipse is $\frac{x^2}{10}+\frac{y^2}{40}=1$.
Reason: $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ is an equation of ellipse if major axis is along y-axis.

• A. Assertion is correct, reason is correct; reason is a correct explanation for assertion.
• B. Assertion is correct, reason is correct; reason is not a correct explanation for assertion
• C. Assertion is correct, reason is incorrect
• D. Assertion is incorrect, reason is correct.

Answer 1

Answer: (A)

Assertion: Since, major axis is along y-axis. Hence, equation of ellipse will be of the form $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1....(i)$
Given that (i) passes through the points $(3,2)$ and $(1,6)$ i.e, they will satisfy it
$\therefore \frac{3^2}{b^2}+\frac{2^2}{a^2}=1\Rightarrow \frac{9}{b^2}+\frac{4}{a^2}=1....(ii)$
and $\frac{1^2}{b^2}+\frac{6^2}{a^2}=1\Rightarrow \frac{1}{b^2}+\frac{36}{a^2}=1....(iii)$
Multiplying (ii) by 9 and then subtracting (iii) from it,
we get $\frac{80}{b^2}=8\Rightarrow b^2=\frac{80}{8}\therefore b^2=10$
from eq (ii), we get $\frac{9}{10}+\frac{4}{a^2}=1\Rightarrow \frac{4}{a^2}=1-\frac{9}{10}\Rightarrow a^2=40$
putting the value of $a^2=40$ and $b^2=10$ in (i), we get $\frac{x^2}{10}+\frac{y^2}{40}=1$