Question

The mid point of the chord \(4x-3y=5\) of the hyperbola \(2x^2-3y^2=12\) is

• A. \(\left(0,-\frac{5}{3}\right)\)
• B. \((2,1)\)
• C. \(\left(\frac{5}{4},0\right)\)
• D. \(\left(\frac{11}{4},2\right)\)

Hint:

For chord midpoint, solve intersection to get a quadratic in \(x\). Midpoint uses \(\frac{x_1+x_2}{2}\), where \(x_1+x_2=-\frac{b}{a}\).

Answer 1

The Correct Option is B

Step 1: Use midpoint of chord concept.
For a conic \(S=0\), the chord given by a line \(L=0\) has midpoint where the line \(L=0\) meets the diameter (line joining midpoints of parallel chords).
Step 2: Solve intersection of hyperbola with line and find midpoint (direct method).
Line:
\[ 4x-3y=5 \Rightarrow y=\frac{4x-5}{3} \]
Substitute into hyperbola:
\[ 2x^2-3\left(\frac{4x-5}{3}\right)^2=12 \]
\[ 2x^2-\frac{(4x-5)^2}{3}=12 \]
Multiply by 3:
\[ 6x^2-(4x-5)^2=36 \]
\[ 6x^2-(16x^2-40x+25)=36 \]
\[ 6x^2-16x^2+40x-25=36 \]
\[ -10x^2+40x-61=0 \Rightarrow 10x^2-40x+61=0 \]
Roots are \(x_1,x_2\). Midpoint x-coordinate:
\[ x_m=\frac{x_1+x_2}{2}=\frac{\frac{40}{10}}{2}=\frac{4}{2}=2 \]
Now find y using line:
\[ y_m=\frac{4(2)-5}{3}=\frac{8-5}{3}=1 \]
Final Answer:
\[ \boxed{(2,1)} \]