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Parabola · Uttarakhand Board (UBSE) STD 11 Science (Uttarakhand - English Medium). 11 questions.

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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 $(y - k)^2 = 4a(x - h)$
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 PM perpendicular from $P(x, y)$ on the directrix
$x + y - 4 = 0$
Then by definition,
$SP = PM$
$\Rightarrow SP^2 = PM^2$^
$\Rightarrow\ (\text{x} - \text{0})^2 + (\text{y} - \text{0})^2 =\Big[\frac{\text{x+y}-4}{\sqrt{1^2+1^2}}\Big]^2$
$\Rightarrow\ \text{x}^2+\text{y}^2=\frac{(\text{x+y}-4)^2}{(\sqrt2)^2}$
$\Rightarrow 2x^2 + 2y^2 = x^2 + y^2 + (-4)^2 + 2xy + 2 \times y \times (-4) + 2 \times (-4) \times x$
$\Rightarrow 2x^2 + 2y^2 = x^2 + y^2 + 16 + 2xy - 8y - 8x$
$\Rightarrow 2x^2 - x^2 + 2y^2 - y^2 - 2xy + 8x + 8y -16 = 0$
$\Rightarrow x^2 + y^2 - 2xy + 8x + 8y - 16 = 0.$
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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 $\frac{\text{y}_2-\text{y}_1}{\text{x}_2-\text{x}_1}=\frac{\text{y}-\text{y}_1}{\text{x}-\text{x}_1}$
$\Rightarrow\ \frac{3-1}{2-(-1)}=\frac{\text{y}-1}{\text{x}-(-1)}  [ \because$ Focus: $(2, 3)$ and vertex: $(-1, 1)]$
$\Rightarrow\ \frac{2}{3}=\frac{\text{y}-1}{\text{x}+1}$
$\Rightarrow 2x + 2 = 3y - 3 $
$\Rightarrow 3y - 2x - 3 - 2 = 0 $
$\Rightarrow 3y - 2x - 5 = 0 ...(i)$
The equation of $\bot$ line to $3y - 2x - 5 = 0$ is$2\text{y} + 3\text{x} + \lambda = 0 ...(\text{ii})$
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 $(x_{1,}y_1).$
$\therefore\ \frac{\text{x}_1+2}{2}=-1\text{ and }\frac{\text{y}_1+3}{2}=1$
$\Rightarrow x_1 + 2 = -2$ and $y_1 + 3 = 2$
$\Rightarrow x_1 = -4$ and $y_1 = -1$
Thus, the directrix meets the axis at $(-4, -1).$
$\therefore$ The prependicular line $2\text{y} + 3\text{x} +\lambda = 0$
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 $2y + 3x + 14 = 0$
Hence, the required equation of directrix is $2y + 3x + 14 = 0.$
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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
$x + y - 3 = 0 ...(i)$
$\Rightarrow y = -x + 3$
$\therefore$ slope of line $= m_1 = -1$
$\therefore$ slope of line $=\ \text{m}_2=\frac{-1}{\text{m}_1}=\frac{-1}{-1}=1$
The equation of line posess through $(-2, 1)$ with slope 1 is
$y - 1 = 1 [x - (-2)] [\because y - y_0 = m(x - x_0)]$
$\Rightarrow y - 1 = x + 2$
$\Rightarrow y - x = 3$
$\Rightarrow y - x - 3 = 0 ...(ii)$
Adding equation $(i)$ and $(ii),$ we get
$2y - 6 = 0$
$\Rightarrow 2y = 6$
$\Rightarrow\ \text{y}=\frac{6}{2}=3$
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 $(x_1, y_1)$ is the mid-point of the line segment joining $(0, 3)$ and $(-2, 1)$
$\therefore\ \text{x}_1=\frac{0-2}{2}\text{ and }\text{y}_1=\frac{3+1}{2}$
$\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 $y^2 = 4ax$
Since, the parabola is passing through the point $(3, 2)$
Hence, it will satisfy the equation of the parabola.
$\therefore 2^2 = 4(a)(3)$
$\Rightarrow\ \text{a}=\frac{1}{3}$
Lenth of the latus ractum is given by,
$4a$
$=\ 4\times\frac{1}{3}$
$=\ \frac{4}{3}$
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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 $y^2 + 6y + 2x + 5 = 0.$
Answer
We have,
$y^2+ 6y = -2x - 5$
$\Rightarrow y^2 + 2 \times y \times 3 + 9 = -2x - 5 + 9$
$\Rightarrow (y + 3)^2 = -2x + 4$
$\Rightarrow (y + 3)^2 = -2(x - 2) ...(i)$
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
$x = x + 2, y = y - 3 ...(ii)$
Using these relation, equation (i) reduces to
$y^2 = -2x ...(iii)$
This is of the form $y^2 = -4ax$. on comparing, we get
$4a = 2$
$\Rightarrow\ \text{a}=\frac{2}{4}=\frac{1}{2}$
$\therefore$ 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
$x^2 - 4x - 8y +12 =0$
$\Rightarrow x^2 - 4x = 8y - 12$
$\Rightarrow x^2 - 2 \times x \times 2 + 4 = 8y - 12 + 4$
$\Rightarrow (x - 2)^2 = 8y - 8$
$\Rightarrow (x - 2)^2 = 8(y - 1) ...(i)$
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
$y^2 = 8y ...(iii)$
This is of the form $x^2 = 4ay.$ on comparing, we get
$4a = 8$
$\Rightarrow a = 2$
$\therefore$ 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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