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3 questions · self-marked practice — reveal the answer and mark yourself.

Question 11 Mark

Find the equation of the parabola whose diretrix is $x+3=0$

Answer

Here equation of diretrixs is $x+a=0$ that is $x+3=0$ comparing we get $a=3$.
$\therefore$ Equation of the parabola $y^2=4 a x$ that is $y^2=12 x$.

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Question 21 Mark

Find the equation of an ellipse whose major axis is on the $\mathrm{X}$-axis and passes through the points $(4,3)$ and $(6,2)$

Answer

Let equation an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,
$a>b$, since major axis is the $\mathrm{X}$-axis.
Also ellipse passes through points $(4,3)$ and $(6,2)$
$
\therefore \frac{(4)^2}{a^2}+\frac{(3)^2}{b^2}=1 \text { and } \frac{(6)^2}{a^2}+\frac{(2)^2}{b^2}=1
$
Solve these equations simultaneous to set $\mathrm{a}^2$ and

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Question 31 Mark

Find the length of the latus rectum of the parabola $y^2=4 a x$ passing through the point $(2$,-6).

Answer

Given equation of the parabola is $y^2=4 a x$ and it passes through point $(2,-6)$.

Substituting $x=2$ and $y=-6$ in $y^2=4 a x$, we get

$\begin{aligned} & \Rightarrow(-6)^2=4 a(2) \\ & \Rightarrow 4 a=18\end{aligned}$

∴ Length of latus rectum = 4a = 18 units

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Maths Answer the following questions in short.

Answer the following questions in short.

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