(Each question 1 marks)

Question 11 Mark

Write the length of the chord of the parabola $y^2 = 4ax$ which passes through the vertex and is inclined to the axis at $\frac{\pi}{4}.$

Answer

Let A be the vertex of the parabola. Then cooridnates of A are (0, 0). Suppose AP is a chord that is inclined to an angle of $\frac{\pi}{4}$ radians to the X axis. Let M be the point where the perpendicular from P intersects the X axis. Let AP = l Then, $\frac{\text{AM}}{\text{l}}=\cos\frac{\pi}{4}$ $\Rightarrow\ \text{AM}=\text{l}\times\frac{1}{\sqrt2}$ $\Rightarrow\ \text{AM}=\frac{\text{l}}{\sqrt2}$ $\text{and }\frac{\text{PM}}{\text{l}}=\sin\frac{\pi}{4}$ $\Rightarrow\ \text{PM}=\text{l}\times\frac{1}{\sqrt2}=\frac{\text{l}}{\sqrt2}$ So, that coordinates of P are $\Big(\frac{\text{l}}{\sqrt2}, \frac{\text{l}}{\sqrt2}\Big).$ Since, P lies on $y^2 = 4ax \therefore\ \Big(\frac{\text{l}}{\sqrt2}\Big)^2=4\text{a}\times\frac{\text{l}}{\sqrt2}$ $\Rightarrow\ \frac{\text{l}^2}{2}=4\text{a}\frac{\text{l}}{\sqrt2}$ $\Rightarrow\ \text{l}=\frac{8\text{a}}{\sqrt2}$ $\Rightarrow\ \text{l}=\frac{4\times\sqrt2\times\sqrt2\text{a}}{\sqrt2}$ $=\ 4\sqrt2\text{a}$ $\Rightarrow\ \text{l}=4\sqrt2\text{a}$

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Question 21 Mark

Write the equation of the parabola whose vertex is at (-3, 0) and the directrix is x + 5 = 0.

Answer

The general equation of the parabola is $(y-k)^2=4 a(x-h)$ Here, the $(h, k)=(-3,0)$ Now, the directrix is given by $x=$ $h-a \Rightarrow-5=-3-a[\because x+5=0 \Rightarrow x=-5] \Rightarrow a=2$ Hence, the equation is given by $(y-0)^2=4(2)(x+3) \Rightarrow y^2=8(x+$ 3)

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Question 31 Mark

Write the equation of the parabola with focus $(0, 0)$ and directrix $x + y - 4 = 0$.

Answer

Let $P(x, y)$ be any point on the parabola whose focus is $S(0,0)$ and the directrix $x+y-4=0$ Draw $P M$ perpendicular from $P(x, y)$ on the directrix $x+y-4=0$ Then by definition, $S P=P M$
$\Rightarrow S P^2=P M^2$
$\Rightarrow(x-0)^2+(y-0)^2=\left[\frac{x+y-4}{\sqrt{1^2+1^2}}\right]^2$
$\Rightarrow x^2+y^2=\frac{(x+y-4)^2}{(\sqrt{2})^2}$
$\Rightarrow 2 x^2+2 y^2=x^2+y^2+(-4)^2+2 x y+2 \times y \times(-4)+2 \times(-4) \times x$
$\Rightarrow 2 x^2+2 y^2=x^2+y^2+16+2 x y-8 y-8 x$
$\Rightarrow 2 x^2-x^2+2 y^2-y^2-2 x y+8 x+8 y-16=0$
$\Rightarrow x^2+y^2-2 x y$
$+8 x+8 y-16=0$

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Question 41 Mark

If the coordinates of the vertex and focus of a parabola are $(-1, 1)$ and $(2, 3)$ respectively, then write the equation of its directrix.

Answer

The equation of line posses through vertex and focus of a parabola is $\frac{y_2-y_1}{x_2-x_1}=\frac{y-y_1}{x-x_1} $
$\Rightarrow \frac{3-1}{2-(-1)}=\frac{y-1}{x-(-1)}[\because$ Focus: $(2,3)$ and vertex: $(-1,1)] $
$\Rightarrow \frac{2}{3}=\frac{y-1}{x+1} $$\Rightarrow 2 x+2=3 y-3 $
$\Rightarrow 3 y-2 x-3-2=0 $
$\Rightarrow 3 y-2 x-5=0 \ldots$ (i) The equation of $\perp$ line to $3 \mathrm{y}-2 \mathrm{x}-5=0$ is $2 \mathrm{y}+3 \mathrm{x}+\lambda=0 \ldots$ (ii)
Let $(x_1, y_1)$ be the coordinates of the point of intersection of the axis and directrix. Then $(-1,1)$ is the mid-point of the line segment joining $(2,3)$ and $(x_1, y_1) \ldots \frac{x_1+2}{2}=-1$ and $\frac{y_1+3}{2}=1 $
$\Rightarrow x_1+2=-2$ and $y_1+3=2 $
$\Rightarrow x_1=-4$ and $y_1$
$=-1$ Thus, the directrix meets the axis at $(-4,-1) \ldots$ The prependicular line $2 \mathrm{y}+3 \mathrm{x}+\lambda=0$ posses through ( $-4,-1$ ). $\therefore 2(-1)+3(-4)+\lambda=0 $
$\Rightarrow-2-12+\lambda=0$
$\Rightarrow \lambda=14$ Putting $\lambda=14$ in equation (ii), we get $2 y+3 x+14=0$ Hence, the required equation of directrix is $2 y+3 x+14=0$.

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Question 51 Mark

Write the coordinates of the vertex of the parabola whose focus is at(-2, 1) and directrix is the line x + y - 3 = 0.

Answer

The equation of directrix is $x+y-3=0 \ldots$ (i) $\Rightarrow y=-x+3 \therefore$ slope of line $=m_1=-1 \therefore$ slope of line $=m_2=\frac{-1}{m_1}=\frac{-1}{-1}=1$ The equation of line posess through $(-2,1)$ with slope 1 is $y-1=1[x-(-2)]\left[\because y-y_0=m(x\right.$ $\left.\left.-x_0\right)\right] \Rightarrow y-1=x+2 \Rightarrow y-x=3 \Rightarrow y-x-3=0 \ldots$ (ii) Adding equation (i) and (ii), we get $2 y-6=0 \Rightarrow 2 y=6$ $\Rightarrow \mathrm{y}=\frac{6}{2}=3$ Putting $y=3$ in equation (i), we get $x+3-3=0 \Rightarrow x=0 \therefore(0,3)$ be the coordinates of the point of intersection of the axis and directrix. Then, coordinates of vertex $\left(x_1, y_1\right)$ is the mid-point of the line segment joining $(0,3)$ and $(-2,1) \therefore \mathrm{x}_1=\frac{0-2}{2}$ and $\mathrm{y}_1=\frac{3+1}{2} \Rightarrow$ required coordinates of vertex are $(-1,2)$.

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Question 61 Mark

If the parabola $y^2 = 4ax$ passes through the point (3, 2), then find the length of its latusrectum.

Answer

We have $y^2=4 a x$ Since, the parabola is passing through the point $(3,2)$ Hence, it will satisfy the equation of the parabola. $\therefore 2^2=4(a)(3) \Rightarrow a=\frac{1}{3}$ Lenth of the latus ractum is given by, $4 a=4 \times \frac{1}{3}=\frac{4}{3}$

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Question 71 Mark

PSQ is a focal chord of the parabola $y^2 = 8x$. If SP = 6, then write SQ.

Answer

The coordinates of the focal chord are $\text{P}(\text{at}^2, 2\text{at}^2)$ and $\text{Q}\Big(\frac{\text{a}}{\text{t}^2},\ \frac{-2\text{a}}{\text{t}}\Big).$ Comparing $y^2 = 8x$ with $y^2 = 4ax$: a = 2 Therefore, the coordinates of the focus S is (2, 0). Given: SP = 6 $\therefore\ \sqrt{(2-2\text{t}^2)^2+(4\text{t})^2=6}$ $\Rightarrow\ \text{t}^4+2\text{t}^2-8=0$ $\Rightarrow\ \text{t}^2=2$ Thus, we have: $\text{SQ}=\sqrt{\Big(2-\frac{2}{\text{t}^2}\Big)^2+\Big(\frac{4}{\text{t}^2}\Big)}=\sqrt{\Big(2-\frac{2}{2}\Big)^2+\Big(\frac{4}{2}\Big)}=3$

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Question 81 Mark

Ifb and care lengths of the segments of any focal chord of the parabola $y^2 = 4ax$, then write the length of its latus-rectum.

Answer

Let S(a, 0) be the focus of the given parabola. Let the end points of the focal chord be $\text{P}(\text{at}^2, 2\text{at})$ and $\text{Q}\Big(\frac{\text{a}}{\text{t}^2},\ \frac{-2\text{a}}{\text{t}}\Big).$ SP and SQ are segments of the focal chord with lengths b and c, respectively. $\therefore$ SP = b, SQ = c Also, $\text{SP}=\sqrt{(\text{a}-\text{at}^2)+4\text{a}^2\text{t}^2}=\text{a}(1+\text{t}^2)$ And, $\text{SQ}=\sqrt{\Big(\text{a}-\frac{\text{a}}{\text{t}^2}\Big)+\frac{4\text{a}^2}{\text{t}^2}}=\text{a}\Big(1+\frac{1}{\text{t}^2}\Big)$ Now, we have: $\frac{1}{\text{SP}}+\frac{1}{\text{SQ}}=\frac{1}{\text{a}(1+\text{t}^2)}+\frac{\text{t}^2}{\text{a}(1+\text{t}^2)}=\frac{1}{\text{a}}$ $\Rightarrow\ \frac{1}{\text{b}}+\frac{1}{\text{c}}=\frac{1}{\text{a}}$ $\Rightarrow\ \frac{\text{b+c}}{\text{bc}}=\frac{1}{\text{a}}$ $\Rightarrow\ \text{a}=\frac{\text{bc}}{\text{b+c}}$ $\therefore$ Length of the latus rectum $=4\text{a}=\frac{4\text{bc}}{\text{b+c}}$

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Question 91 Mark

Write the axis of symmetry of the parabola $y^2=x$.

Answer

The axis of symmetry of the parabola $y^2=x$ is x-axis.

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Question 101 Mark

Write the distance between the vertex and focus of the parabola $y^2+6 y+2 x+5=0$

Answer

We have, $y^2+6 y=-2 x-5 \Rightarrow y^2+2 \times y \times 3+9=-2 x-5+9 \Rightarrow(y+3)^2=-2 x+4 \Rightarrow(y+3)^2=-2(x-2) \ldots$ (i) Shifting the origin to point $(2,-3)$ without rotating the axes and denoting the new coordinates w.r.t these axes by x and y . we have $x=x+2, y=y-3 \ldots$...(ii) Using these relation, equation (i) reduces to $y^2=-2 x$...(iii) This is of the form $y^2=-4 a x$. on comparing, we get $4 \mathrm{a}=2 \Rightarrow \mathrm{a}=\frac{2}{4}=\frac{1}{2} \therefore$ The distance between the vertex and focus of the parabola is $\frac{1}{2}$.

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Question 111 Mark

Write the equation of the directrix of the parabola $x^2 - 4x - 8y + 12 = 0$.

Answer

The given system of equation is $x^2-4 x-8 y+12=0 \Rightarrow x^2-4 x=8 y-12 \Rightarrow x^2-2 \times x \times 2+4=8 y-12+4 \Rightarrow(x-2)^2=$ $8 y-8 \Rightarrow(x-2)^2=8(y-1) \ldots$ (i) Shifting the origin to point $(2,1)$ without rotating the axes and denoting the new coordinates w.r.t these axes by $x$ and $y$, we have $x=x+2, y=y-3$...(ii) Using these relation, equation (i) reduces to $y^2=8 y \ldots$ (iii) This is of the form $x^2=4 a y$. on comparing, we get $4 a=8 \Rightarrow a=2 \therefore$ equation of the directrix of the parabola w.r.t new axes is $y=-2 \therefore y=-2+1$ [Using equation (ii)] $\Rightarrow y=-1 \Rightarrow y+1=0 \therefore$ equation of the directrix of the parabola w.r.t old axes is $y+1=0$.

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