Question

If the line \(3x-my+5=0\) is a tangent to the hyperbola \(3x^2-4y^2=300\), then the square of the \(Y\)-intercept made by this tangent line is

• A. \(\dfrac{25}{3}\)
• B. \(\dfrac{35}{3}\)
• C. \(\dfrac{45}{7}\)
• D. \(\dfrac{15}{7}\)

Hint:

For the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] the tangent with slope \(m\) is \[ y=mx\pm \sqrt{a^2m^2-b^2}. \] This formula is very useful in tangent-related problems.

Answer 1

The Correct Option is D

Step 1: Rewrite the hyperbola in standard form.
We are given $ 3x^2 - 4y^2 = 300 $. Divide both sides by 300: \[ \frac{x^2}{100} - \frac{y^2}{75} = 1 \] So $ a^2 = 100 $ and $ b^2 = 75 $. This is a horizontal hyperbola.
Step 2: Substitute the line into the hyperbola directly.
From $ 3x - my + 5 = 0 $, we get $ x = \frac{my-5}{3} $. Substitute into the hyperbola equation to find conditions for tangency without using the slope-intercept tangent formula.
Step 3: Derive the quadratic in y.
\[ \frac{(my-5)^2}{900} - \frac{y^2}{75} = 1 \implies (m^2-12)y^2 - 10my - 875 = 0 \]
Step 4: Apply the discriminant condition for tangency.
For a tangent, exactly one intersection means discriminant = 0: \[ (-10m)^2 + 4(m^2-12)(875) = 0 \] \[ 100m^2 + 3500m^2 - 42000 = 0 \implies m^2 = \frac{35}{3} \]
Step 5: Find the y-intercept of the line.
At $ x = 0 $: $ -my + 5 = 0 \implies y = \frac{5}{m} $. Square: $ \frac{25}{m^2} $.
Step 6: Substitute $ m^2 $ and compute the answer.
\[ \frac{25}{m^2} = \frac{25}{\frac{35}{3}} = \frac{75}{35} = \frac{15}{7} \] \[ \boxed{\dfrac{15}{7}} \]