Question

Read the following statements carefully and choose the correct option(s)
$\text{(I)}\frac{x^2}{25}+\frac{y^2}{16}=1$ and $12x^2-4y^2=27$ intersect orthogonally.

$\text{(II)}$Locus of vertex of a parabola with focus at $(2,3)$ and length of latus rectum is $8$, is a circle.

$\text{(III)}$The two circles $x^2+y^2-10x+4y-20=0$ and $x^2+y^2+14x-6y+22=0$ intersect each other.

• A. $\text{(I)}$ is correct
• B. $\text{(II)}$ is correct
• C. $\text{(III)}$ is incorrect
• D. $\text{(III)}$ is correct

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\text{(I)}$ is correct
B $\text{(II)}$ is correct
C $\text{(III)}$ is incorrect

$\text{(I)}$
Given equation are $\frac{x^2}{25}+\frac{y^2}{16}=1$ and $12x^2-4y^2=27$
Now, $\frac{x^2}{25}+\frac{y^2}{16}=1$
Eccentricity is
$e=\sqrt{1-\frac{b^2}{a^2}}\Rightarrow e=\sqrt{1-\frac{16}{25}}=\frac{3}{5}$
Foci $=(\pm ae,0)=(\pm 3,0)$

For $12x^2-4y^2=27,$
$\frac{x^2}{\left(\frac{27}{12}\right)}-\frac{y^2}{\left(\frac{27}{4}\right)}=1$
Eccentricity is
$e^{\prime }=\sqrt{1+\frac{b^2}{a^2}}\Rightarrow e^{\prime }=\sqrt{\frac{\frac{27}{12}+\frac{27}{4}}{\frac{27}{12}}}\Rightarrow e^{\prime }=2$
Foci $=(\pm a^{\prime }{e}^{\prime },0)=(\pm 3,0)$
As both the curves have same focii, so they intersect orthogonally.
$\text{(I)}$ is correct.

$\text{(II)}$
Let vertex be $P(x,y)$
Focus $=(2,3)$
Distance between focus and vertex is one fourth of length of latus rectum, so
$\sqrt{(x-2{)}^2+(y-3{)}^2}=\frac{8}{4}\Rightarrow (x-2{)}^2+(y-3{)}^2=4$
which is a circle.
$\text{(II)}$ is correct.

$\text{(III)}$
Given circles are $x^2+y^2-10x+4y-20=0$ and $x^2+y^2+14x-6y+22=0$
Centre and radius of the given circles are
$C_1=(5,-2),r_1=7C_2=(-7,3),r_2=6$
Now, distance between centres is
$C_1{C}_2=\sqrt{{12}^2+5^2}=13r_1+r_2=13$
So, the two circles touch each other
$\text{(III)}$ is incorrect.