WBJEE · Maths · Ellipse
For the variable, the locus of the point of intersection of the lines \(3 t x-2 y+6 t=0\) and \(3 x+2 t y-6=0\) is
- A the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\)
- B the ellipse \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\)
- C the hyperbola \(\frac{x^{2}}{4}-\frac{y^{2}}{9}=1\)
- D the hyperbola \(\frac{x^{2}}{9}-\frac{y^{2}}{4}=1\)
Answer & Solution
Correct Answer
(A) the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\)
Step-by-step Solution
Detailed explanation
Given equation of lines are \[ 3 t \times-2 y+6 t=0 \] and \(\quad 3 x+2 t y-6=0\) On multiplying Eq. (i) by \(t\) and then adding in Eq. (ii), we get \(\left(3 t^{2}+3\right) x+6 t^{2}-6=0\) \(\Rightarrow \quad x=\frac{2\left(1-t^{2}\right)}{\left(1+t^{2}\right)}\)…
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