Question:
A normal is drawn to the hyperbola \( 9x^{2}-16y^{2}=144 \) at one of the ends of its latus rectum. If that end lies in the third quadrant and the equation of the normal is \( ax+by+c=0 \) then \( \frac{b+c}{a} = \)
A normal is drawn to the hyperbola \( 9x^{2}-16y^{2}=144 \) at one of the ends of its latus rectum. If that end lies in the third quadrant and the equation of the normal is \( ax+by+c=0 \) then \( \frac{b+c}{a} = \)
Show Hint
Be extra careful with signs when substituting points in the third quadrant. Both \( x_1 \) and \( y_1 \) are negative, which changes the signs of the terms in the formula to positive when shifted to one side.
Updated On: Jun 8, 2026
- \( \frac{44}{25} \)
- \( \frac{84}{25} \)
- \( \frac{55}{16} \)
- \( \frac{145}{16} \)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Concept:
First, convert the hyperbola into standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
The coordinates of the ends of the latus rectum are \( (\pm ae, \pm \frac{b^2}{a}) \). For the third quadrant, both coordinates must be negative: \( (-ae, -\frac{b^2}{a}) \).
The equation of a normal to a hyperbola at a specific point \( (x_1, y_1) \) is:
\[
\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2
\]
Step 1: Converting to standard form and finding parameters.
Divide \( 9x^2 - 16y^2 = 144 \) by 144: \[ \frac{x^2}{16} - \frac{y^2}{9} = 1 \implies a^2 = 16, \quad b^2 = 9 \] Calculating focal parameters: \[ ae = \sqrt{a^2 + b^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \] The third-quadrant end of the latus rectum is: \[ x_1 = -ae = -5, \quad y_1 = -\frac{b^2}{a} = -\frac{9}{4} \]
Step 2: Writing the equation of the normal line.
Substitute \( x_1 = -5 \) and \( y_1 = -\frac{9}{4} \) into the normal line formula: \[ \frac{16x}{-5} + \frac{9y}{-\frac{9}{4}} = 16 + 9 \] \[ -\frac{16}{5}x - 4y = 25 \] Multiply the entire equation by -5 to clean up the coefficients: \[ 16x + 20y + 125 = 0 \] Comparing this with \( ax + by + c = 0 \), we find \( a = 16 \), \( b = 20 \), and \( c = 125 \).
Step 3: Calculating the final value.
\[ \frac{b + c}{a} = \frac{20 + 125}{16} = \frac{145}{16} \] This matches Option (D) perfectly.
Step 1: Converting to standard form and finding parameters.
Divide \( 9x^2 - 16y^2 = 144 \) by 144: \[ \frac{x^2}{16} - \frac{y^2}{9} = 1 \implies a^2 = 16, \quad b^2 = 9 \] Calculating focal parameters: \[ ae = \sqrt{a^2 + b^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \] The third-quadrant end of the latus rectum is: \[ x_1 = -ae = -5, \quad y_1 = -\frac{b^2}{a} = -\frac{9}{4} \]
Step 2: Writing the equation of the normal line.
Substitute \( x_1 = -5 \) and \( y_1 = -\frac{9}{4} \) into the normal line formula: \[ \frac{16x}{-5} + \frac{9y}{-\frac{9}{4}} = 16 + 9 \] \[ -\frac{16}{5}x - 4y = 25 \] Multiply the entire equation by -5 to clean up the coefficients: \[ 16x + 20y + 125 = 0 \] Comparing this with \( ax + by + c = 0 \), we find \( a = 16 \), \( b = 20 \), and \( c = 125 \).
Step 3: Calculating the final value.
\[ \frac{b + c}{a} = \frac{20 + 125}{16} = \frac{145}{16} \] This matches Option (D) perfectly.
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 sections of a cone Questions
- The equation of the parabola with focus at $(3, 0)$ and directrix $x + 3 = 0$ is:
- If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12, then the length of its latus rectum is:
- The line $y = mx + c$ touches the ellipse $9x^2 + 16y^2 = 144$ if the value of $c^2$ is:
- If the eccentricity of a hyperbola is $\sqrt{3}$, then the eccentricity of its conjugate hyperbola is:
- The equation of the tangent to the parabola $y^2 = 8x$ which is parallel to the line $2x - y + 5 = 0$ is:
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