Qus : 1 2 If the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ are coincide, then the value of $b^2$
1 25 2 16 3 64 4 49 Go to Discussion
Solution Qus : 2 3 At how many points the following curves intersect $\frac{{y}^2}{9}-\frac{{x}^2}{16}=1$ and $\frac{{x}^2}{4}+\frac{{(y-4)}^2}{16}=1$
1 0 2 1 3 2 4 4 Go to Discussion
Solution Qus : 3 3 If ${\Bigg{(}\frac{x}{a}\Bigg{)}}^2+{\Bigg{(}\frac{y}{b}\Bigg{)}}^2=1$, $(a{\gt}b)$ and ${x}^2-{y}^2={c}^2$ cut at right angles, then
1 $${a}^2+{b}^2=2{c}^2$$ 2 $${b}^2-{a}^2=2{c}^2$$ 3 $${a}^2-{b}^2=2{c}^2$$ 4 $${a}^2-{b}^2={c}^2$$ Go to Discussion Solution If $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$ are orthogonal.
Then
$a^2-b^2=c^2-d^2$
Similarly
If $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $x^2-y^2=c^2$ are orthogonal.
It means
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{c^2}+\frac{y^2}{-c^2}=1$ are orthogonal
Then
$a^2-b^2=c^2-(-c^2)$
$a^2-b^2=2c^2$
Qus : 4 1 If the line $a^2 x + ay +1=0$, for some real number $a$, is normal to the curve $xy=1$
then
1 $a<0$ 2 $$0<a<1$$ 3 $$a>0$$ 4 $$-1<a<1$$ Go to Discussion Solution
Problem:
The line \[
a^2x + ay + 1 = 0
\] is normal to the curve \[
xy = 1
\]. Find possible values of \( a \in \mathbb{R} \).
Step 1: Slope of Line
Rewrite: \[
y = -a x - \frac{1}{a}
\] → slope = \( -a \)
Step 2: Curve Derivative
\[
xy = 1 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}
\]
Slope of normal = \( \frac{x}{y} \)
Match Slopes
\[
-a = \frac{x}{y} \Rightarrow x = -a y
\]
Plug into Curve
\[
xy = 1 \Rightarrow (-a y)(y) = 1 \Rightarrow y^2 = -\frac{1}{a}
\]
For real \( y \), we need \( a < 0 \)
✅ Final Answer:
\[
\boxed{a < 0}
\]
Qus : 5 2 Equation of the common tangents with a positive slope to the circle
and
is
1 2 3 4 Go to Discussion
Solution Qus : 7 3 The locus of the intersection of the two lines $\sqrt{3} x-y=4k\sqrt{3}$ and $k(\sqrt{3}x+y)=4\sqrt{3}$, for different
values of k, is a hyperbola. The eccentricity of the hyperbola is:
1 $1.5$ 2 $\sqrt{3}$ 3 $2$ 4 $\frac{\sqrt{3}}{2}$ Go to Discussion
Solution Qus : 8 3 If the foci of the ellipse $b^{2}x^{2}+16y^{2}=16b^{2}$ and the hyperbola $81x^{2}-144y^{2}=\frac{81 \times 144}{25}$ coincide, then the value of $b$, is
1 $1$ 2 $\sqrt{5}$ 3 $\sqrt{7}$ 4 $3$ Go to Discussion
Solution Qus : 9 1 If $3x + 4y + k = 0$ is a tangent to the hyperbola ,$9x^{2}-16y^{2}=144$ then the value of $K$ is
1 0 2 1 3 -1 4 -3 Go to Discussion
Solution Qus : 10 2 If PQ is a double ordinate of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ such that OPQ is an equilateral triangle,
where O is the centre of the hyperbola, then which of the following is true?
1 $b^{2}>\frac{-a^{2}}{\sqrt{3}}$ 2 $b^{2}>\frac{a^{2}}{3}$ 3 $b^{2}<\frac{a^{2}}{3}$ 4 $b^{2}<\frac{-a^{2}}{3}$ Go to Discussion
Solution Qus : 11 2
Find foci of the equation $x^2 + 2x – 4y^2 + 8y – 7 = 0$
1 $$(\sqrt[]{5}\pm1,1)$$ 2 $$(-1\pm\sqrt[]{5},1)$$ 3 $$(-1,\sqrt[]{5}\pm1)$$ 4 $$(1,-1\pm\sqrt[]{5})$$ Go to Discussion Solution
Finding Foci of a Conic
Given Equation: \( x^2 + 2x - 4y^2 + 8y - 7 = 0 \)
Step 1: Complete the square
⇒ \( (x + 1)^2 - 4(y - 1)^2 = 4 \)
Rewriting: \( \frac{(x + 1)^2}{4} - \frac{(y - 1)^2}{1} = 1 \)
This is a horizontal hyperbola with:
Center: \( (-1, 1) \)
\( a^2 = 4 \), \( b^2 = 1 \)
\( c = \sqrt{a^2 + b^2} = \sqrt{5} \)
✅ Foci:
\( (-1 \pm \sqrt{5},\ 1) \)
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and 
is