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
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
Show 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.
Updated On: Jun 25, 2026
- \(\dfrac{25}{3}\)
- \(\dfrac{35}{3}\)
- \(\dfrac{45}{7}\)
- \(\dfrac{15}{7}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Write the hyperbola in standard form.
Given hyperbola: \[ 3x^2-4y^2=300 \] Dividing by \(300\), \[ \frac{x^2}{100}-\frac{y^2}{75}=1 \] Thus, \[ a^2=100,\quad b^2=75 \]
Step 2: Write the tangent in slope form.
Given line: \[ 3x-my+5=0 \] Rearranging, \[ my=3x+5 \] \[ y=\frac{3}{m}x+\frac{5}{m} \] Hence, \[ \text{slope}=\frac{3}{m} \] and \(Y\)-intercept is \[ \frac{5}{m} \]
Step 3: Use tangent condition for hyperbola.
For the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] the tangent with slope \(k\) is \[ y=kx\pm \sqrt{a^2k^2-b^2} \] Here, \[ k=\frac{3}{m} \] So, \[ y=\frac{3}{m}x\pm \sqrt{100\left(\frac{3}{m}\right)^2-75} \] Comparing with \[ y=\frac{3}{m}x+\frac{5}{m}, \] we get \[ \frac{5}{m}=\pm \sqrt{\frac{900}{m^2}-75} \] Squaring both sides, \[ \frac{25}{m^2}=\frac{900}{m^2}-75 \] Multiplying by \(m^2\), \[ 25=900-75m^2 \] \[ 75m^2=875 \] \[ m^2=\frac{35}{3} \]
Step 4: Find the square of the \(Y\)-intercept.
\(Y\)-intercept is \[ \frac{5}{m} \] Therefore, its square is \[ \left(\frac{5}{m}\right)^2 = \frac{25}{m^2} \] Substitute \[ m^2=\frac{35}{3} \] \[ \frac{25}{m^2} = \frac{25}{35/3} \] \[ = \frac{25\times 3}{35} \] \[ = \frac{15}{7} \]
Step 5: Final conclusion.
Hence, the square of the \(Y\)-intercept is \[ \boxed{\frac{15}{7}} \]
Given hyperbola: \[ 3x^2-4y^2=300 \] Dividing by \(300\), \[ \frac{x^2}{100}-\frac{y^2}{75}=1 \] Thus, \[ a^2=100,\quad b^2=75 \]
Step 2: Write the tangent in slope form.
Given line: \[ 3x-my+5=0 \] Rearranging, \[ my=3x+5 \] \[ y=\frac{3}{m}x+\frac{5}{m} \] Hence, \[ \text{slope}=\frac{3}{m} \] and \(Y\)-intercept is \[ \frac{5}{m} \]
Step 3: Use tangent condition for hyperbola.
For the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] the tangent with slope \(k\) is \[ y=kx\pm \sqrt{a^2k^2-b^2} \] Here, \[ k=\frac{3}{m} \] So, \[ y=\frac{3}{m}x\pm \sqrt{100\left(\frac{3}{m}\right)^2-75} \] Comparing with \[ y=\frac{3}{m}x+\frac{5}{m}, \] we get \[ \frac{5}{m}=\pm \sqrt{\frac{900}{m^2}-75} \] Squaring both sides, \[ \frac{25}{m^2}=\frac{900}{m^2}-75 \] Multiplying by \(m^2\), \[ 25=900-75m^2 \] \[ 75m^2=875 \] \[ m^2=\frac{35}{3} \]
Step 4: Find the square of the \(Y\)-intercept.
\(Y\)-intercept is \[ \frac{5}{m} \] Therefore, its square is \[ \left(\frac{5}{m}\right)^2 = \frac{25}{m^2} \] Substitute \[ m^2=\frac{35}{3} \] \[ \frac{25}{m^2} = \frac{25}{35/3} \] \[ = \frac{25\times 3}{35} \] \[ = \frac{15}{7} \]
Step 5: Final conclusion.
Hence, the square of the \(Y\)-intercept is \[ \boxed{\frac{15}{7}} \]
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