Question
Eliminate $\theta$ from given equations.
$ x=a \cot \theta-b \operatorname{cosec} \theta$
$y=a \cot \theta+b \operatorname{cosec} \theta $
$ x=a \cot \theta-b \operatorname{cosec} \theta$
$y=a \cot \theta+b \operatorname{cosec} \theta $
✓
Answer
$ x=a \cot \theta-b \operatorname{cosec} \theta........1$
$y=a \cot \theta+b \operatorname{cosec} \theta........2 $
Adding equations (I) and (II).
$ x+y=2 a \cot \theta$
$\therefore \cot \theta=\frac{x+y}{2 a} $
Subtracting equation (II) from (I),
$ y-x=2 b \operatorname{cosec} \theta$
$\therefore \operatorname{cosec} \theta=\frac{y-x}{2 b} $
Now, $\operatorname{cosec}^2 \theta-\cot ^2 \theta=1$
$ \therefore\left(\frac{y-x}{2 b}\right)^2-\left(\frac{y+x}{2 a}\right)^2=1$
$\therefore \frac{(y-x)^2}{4 b^2}-\frac{(y+x)^2}{4 a^2}=1$
$\text { or }\left(\frac{y-x}{b}\right)^2-\left(\frac{y+x}{a}\right)^2=4 $
$y=a \cot \theta+b \operatorname{cosec} \theta........2 $
Adding equations (I) and (II).
$ x+y=2 a \cot \theta$
$\therefore \cot \theta=\frac{x+y}{2 a} $
Subtracting equation (II) from (I),
$ y-x=2 b \operatorname{cosec} \theta$
$\therefore \operatorname{cosec} \theta=\frac{y-x}{2 b} $
Now, $\operatorname{cosec}^2 \theta-\cot ^2 \theta=1$
$ \therefore\left(\frac{y-x}{2 b}\right)^2-\left(\frac{y+x}{2 a}\right)^2=1$
$\therefore \frac{(y-x)^2}{4 b^2}-\frac{(y+x)^2}{4 a^2}=1$
$\text { or }\left(\frac{y-x}{b}\right)^2-\left(\frac{y+x}{a}\right)^2=4 $
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$\sqrt3\text{x}^2+10\text{x}-8\sqrt3=0$
→$\sqrt3\text{x}^2+10\text{x}-8\sqrt3=0$
Solve the following simultaneous equations.
$\frac{2}{x}+\frac{2}{3 y}=\frac{1}{6} ; \frac{3}{x}+\frac{2}{y}=0$
→$\frac{2}{x}+\frac{2}{3 y}=\frac{1}{6} ; \frac{3}{x}+\frac{2}{y}=0$
In what ratio does the point $(1,3)$ divide line segment joining the points $(3,6)$ and $(-5,-6)$ ?
→In $\triangle\text{ABC},$ P divides the side AB such that AP : PB = 1 : 2. Q is a point in AC such that PQ || BC. Find the ratio of the areas of $\triangle\text{APQ},$ and trapezium BPQC.
→In the given figure, O is the centre of the circle and BCD is tangent to it at C. Prove that $\angle\text{BAC}+\angle\text{ACD}=90^{\circ}.$
→Find the sum of the following arithmetic series:
$7+10\frac{1}{2}+14+...+84$
→$7+10\frac{1}{2}+14+...+84$
Chords AB and CD of a circle intersect in point Q in the interior of a circle of as shown in the figure. If m(arc AD) = 20°, and m(arc BC) = 36°, then find $\angle$BQC.
→Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients:
$2\sqrt3\text{x}^2-\text{5x}+\sqrt3$
→$2\sqrt3\text{x}^2-\text{5x}+\sqrt3$
Express the following as a fraction in simplest form:
$2.\bar4$
→$2.\bar4$
Write the value of $k$ for which the quadratic equation $x^2 - kx + 4 = 0$ has equal roots.
→