2 Marks Questions

Question 12 Marks

Write the coordinates of the centre of the hyperbola $\frac{(x-1)^2}{16}-\frac{(y+2)^2}{9}=1$

Answer

We know that :
Centre of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is at $(0,0)$.
$\therefore \quad x-1=0 \quad \therefore x=1$
Similarly
$y+2=0$
$\therefore y=-2$
So, coordinates of centre $=(1,-2)$

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Question 22 Marks

If a double ordinate of the parabola $y^2=4 a x$ be of length $8 a$, then prove that the angle between the lines joining the vertex of the parabola to the ends of this double ordinate is right angle.

Answer

Given parabola is
\[
y^2=4 a x
\]

We know that, the chord perpendicular to the axis of parabola is called double ordinate.
Let point B $(h, 4 a)$ and $A (h,-4 a)$.
Point B will satisfly equation of parabola.
So,
\[
\begin{aligned}
y^2 & =4 a x \\
(4 a)^2 & =4 a \times h \\
h & =4 a
\end{aligned}
\]
So, here B $(4 a, 4 a)$ and $A (4 a,-4 a)$.
Here we need to prove that $\angle BOA =90^{\circ}$
Slope of line OB $\left(m_1\right)=\frac{y_2-y_1}{x_2-x_1}=\frac{4 a-0}{4 a-0}=1$
Similarly, slope of line OA
\[
\begin{aligned}
\left(m_2\right) & =\frac{y_2-y_1}{x_2-x_1}=-\frac{4 a-0}{4 a}=-1 \\
\therefore \quad m_1 \times m_2 & =-1
\end{aligned}
\]
$\therefore$ The lines joining the ordinates to the origin are at right angle.

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Question 32 Marks

For what point of the parabola $y^2=18 x$ is the ordinate equal to three times abscissa?

Answer

Let us assume that abscissa of point $P =x_1$
$\therefore$ The ordinate of the point P is $3 x_1$
So coordinates of point at parabola $=\left(x_1, 3 x_1\right)$
Above point will satisfy the equation of parabola.
$\therefore \quad\left(3 x_1\right)^2=18 \times x_1$
$9 x_1^2=18 x_1$
$\Rightarrow \quad \begin{aligned} & & x_1{ }^2 & =2 x_1\end{aligned}$
$x_1^2-2 x_1=0$
$\Rightarrow \quad x_1\left(x_1-2\right)=0$
$\therefore \quad x_1=0,2$
$\therefore \quad$ Ordinate $=3 x_1$
$=3 \times 2=6$
Hence, coordinates of point on parabola $=(2,6)$

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Question 42 Marks

An equilatral triangle is inscribed in the parabola $y^2=8 x$, with one of its vertex is the vertex of the parabola. Calculate the length of the side of that triangle.

Answer

Let the side of equilatral triangle inscribed in the parabola be P .

Now in $\triangle OAC$,
\[
\begin{array}{rlrl}
OA^2 & =OC^2+AC^2 \\
\Rightarrow & P^2 & =OC^2+\left(\frac{P}{2}\right)^2 \quad \because AC=\frac{1}{2} AB \\
\Rightarrow \quad & P^2 & =OC^2+\frac{P^2}{4} \\
\Rightarrow \quad & OC^2 & =P^2-\frac{P^2}{4}=\frac{3 P^2}{4} \therefore OC=\frac{\sqrt{3}}{2} P
\end{array}
\]
$\therefore$ Coordinates of $A \left(\frac{\sqrt{3}}{2} P , \frac{ P }{2}\right)$, because it is at parabola.
\[
\begin{array}{rlrl}
\therefore & & \left(\frac{P}{2}\right)^2 & =8 \times \frac{\sqrt{3}}{2} P \\
\Rightarrow & \frac{P^2}{4} & =4 \sqrt{3} P \\
\Rightarrow & & P^2=16 \sqrt{3} P \\
\Rightarrow & P & =16 \sqrt{3}
\end{array}
\]
So, side of equilateral triangle $=16 \sqrt{3}$

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Question 52 Marks

If equation of parabola $y^2=8 x$, then find the coordinates of focus, axis, equation of directrix and length of latus rectum.

Answer

Clearly axis of parabola $y^2=8 x$ is $x$-axis.

Since cofficient of $x$ in given equation is positive, so parabola opens towards right. On comparing the given equation with $y^2=4 a x$,
$
\Rightarrow \quad a=2
$
So focus of parabola is $(2,0)$ and equation of directrix is $x=-2$
Length of latus rectum $4 a=4 \times 2=8$.

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Question 62 Marks

Write the equation of the axis of the parabola $9 y^2$ $-16 x-12 y-57=0$.

Answer

Given,
$9 y^2-16 x-12 y-57=0$
$\begin{array}{ll}\Rightarrow & 9 y^2-12 y=16 x+57 \\ \Rightarrow & 9 y^2-12 y+4=16 x+57+4\end{array}$
$\begin{array}{l}\Rightarrow \\ \text { Let } 3 y-2= Y \text { and } 16 x+61= X \end{array}$
$\Rightarrow \quad Y^2=X$
Equation of axis$Y=0$
$\therefore \quad 3 y-2=0$
or$\quad$$\quad$$3 y=2$

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