Question:
The locus of midpoints of points of intersection of
\[
x\cos\theta+y\sin\theta=1
\]
with the coordinate axes is
The locus of midpoints of points of intersection of
\[
x\cos\theta+y\sin\theta=1
\]
with the coordinate axes is
Show Hint
For a line intersecting coordinate axes, first find the intercepts by putting \(y=0\) and \(x=0\), then use the midpoint formula.
Updated On: Jun 22, 2026
- \(x^2+y^2=4\)
- \(\dfrac{1}{x^2}+\dfrac{1}{y^2}=4\)
- \(\dfrac{1}{x^2}+\dfrac{1}{y^2}=\dfrac12\)
- \(x^2+y^2=2\)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Find intercepts on coordinate axes.
Given line: \[ x\cos\theta+y\sin\theta=1 \] On \(x\)-axis, \(y=0\): \[ x\cos\theta=1 \] \[ x=\sec\theta \] So, point is \[ (\sec\theta,0) \] On \(y\)-axis, \(x=0\): \[ y\sin\theta=1 \] \[ y=\cosec\theta \] So, point is \[ (0,\cosec\theta) \]
Step 2: Let midpoint be \((x,y)\).
\[ x=\frac{\sec\theta}{2} \] and \[ y=\frac{\cosec\theta}{2} \]
Step 3: Express \(\cos\theta\) and \(\sin\theta\).
From \[ x=\frac{\sec\theta}{2}, \] we get \[ \sec\theta=2x \] So, \[ \cos\theta=\frac{1}{2x} \] Similarly, \[ \cosec\theta=2y \] So, \[ \sin\theta=\frac{1}{2y} \]
Step 4: Use identity.
\[ \sin^2\theta+\cos^2\theta=1 \] Substituting, \[ \left(\frac{1}{2y}\right)^2+\left(\frac{1}{2x}\right)^2=1 \] \[ \frac{1}{4y^2}+\frac{1}{4x^2}=1 \]
Step 5: Simplify.
Multiplying by \(4\), \[ \frac{1}{x^2}+\frac{1}{y^2}=4 \]
Step 6: Final conclusion.
Therefore, \[ \boxed{\frac{1}{x^2}+\frac{1}{y^2}=4} \]
Given line: \[ x\cos\theta+y\sin\theta=1 \] On \(x\)-axis, \(y=0\): \[ x\cos\theta=1 \] \[ x=\sec\theta \] So, point is \[ (\sec\theta,0) \] On \(y\)-axis, \(x=0\): \[ y\sin\theta=1 \] \[ y=\cosec\theta \] So, point is \[ (0,\cosec\theta) \]
Step 2: Let midpoint be \((x,y)\).
\[ x=\frac{\sec\theta}{2} \] and \[ y=\frac{\cosec\theta}{2} \]
Step 3: Express \(\cos\theta\) and \(\sin\theta\).
From \[ x=\frac{\sec\theta}{2}, \] we get \[ \sec\theta=2x \] So, \[ \cos\theta=\frac{1}{2x} \] Similarly, \[ \cosec\theta=2y \] So, \[ \sin\theta=\frac{1}{2y} \]
Step 4: Use identity.
\[ \sin^2\theta+\cos^2\theta=1 \] Substituting, \[ \left(\frac{1}{2y}\right)^2+\left(\frac{1}{2x}\right)^2=1 \] \[ \frac{1}{4y^2}+\frac{1}{4x^2}=1 \]
Step 5: Simplify.
Multiplying by \(4\), \[ \frac{1}{x^2}+\frac{1}{y^2}=4 \]
Step 6: Final conclusion.
Therefore, \[ \boxed{\frac{1}{x^2}+\frac{1}{y^2}=4} \]
Was this answer helpful?
0
0
Top AP EAPCET Mathematics Questions
- $D = \left\{x\in\mathbb{R}: f\left(x\right) =\sqrt{\frac{x - \left|x\right|}{x - \left[x\right]}} \text{is defined} \right\}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4 + x^{2}}$. Then $D \cap C$
- $f\left(x\right) = \frac{x}{e^{x}-1} + \frac{x}{2} + 2 \cos^{3} \frac{x}{2} $ on $R-\left\{0\right\} $ is
- Consider the following lists.
List-I List-II (A) $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ (I) $[\frac{1}{3} , 1 ]$ (B) $(x)=|[x]|,x \in [R$ (II) Z (C) $h(x) = |x - [x]| , x \in [R$ (III) W (D) $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ (IV) [0, 1) (V) { -1, 1} - If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$. then the system of three equations $2x + 3 | y | + 5[z] = 0, x + |y| - 2[z] = 4, x + |y| + |z| = 1$ has
- Investigate the values of $\lambda$ and $\mu$ for the system $x+2 y+3 z=6, x+3 y+5 z=9$, $2 x+5 y+\lambda z=\mu$ and match the values in List - I with the items in List - II.
List I List II (A) $\lambda=8, \mu \neq 15$ 1. Infinitely many solutions (B) $\lambda \neq 8, \mu \in R$ 2. No solution (C) $\lambda=8, \mu=15$ 3. Unique solution
View More Questions
Top AP EAPCET Coordinate Geometry Questions
- The perimeter of the locus of the point \( P \) which divides the line segment \( QA \) internally in the ratio 1:2, where \( A = (4, 4) \) and \( Q \) lies on the circle \( x^2 + y^2 = 9 \), is:
- P is a point on \( x + y + 5 = 0 \), whose perpendicular distance from \( 2x + 3y + 3 = 0 \) is \( \sqrt{13} \), then the coordinates of P are:
A line \( L \) intersects the lines \( 3x - 2y - 1 = 0 \) and \( x + 2y + 1 = 0 \) at the points \( A \) and \( B \). If the point \( (1,2) \) bisects the line segment \( AB \) and \( \frac{a}{b} x + \frac{b}{a} y = 1 \) is the equation of the line \( L \), then \( a + 2b + 1 = ? \)
A line \( L \) passing through the point \( (2,0) \) makes an angle \( 60^\circ \) with the line \( 2x - y + 3 = 0 \). If \( L \) makes an acute angle with the positive X-axis in the anticlockwise direction, then the Y-intercept of the line \( L \) is?
If the slope of one line of the pair of lines \( 2x^2 + hxy + 6y^2 = 0 \) is thrice the slope of the other line, then \( h \) = ?
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions