(A) Given equations of lines are:$3tx - 2y + 6t = 0$ $(i)$$3x + 2ty - 6 = 0$ $(ii)$From $(i)$, $3t(x+2) = 2y \Rightarrow t = \frac{2y}{3(x+2)}$.Subs

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the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$

(A) Given equations of lines are:$3tx - 2y + 6t = 0$ $(i)$$3x + 2ty - 6 = 0$ $(ii)$From $(i)$, $3t(x+2) = 2y \Rightarrow t = \frac{2y}{3(x+2)}$.Substitute $t$ into $(ii)$:$3x + 2\left(\frac{2y}{3(x+2)}\right)y - 6 = 0$$3x(3(x+2)) + 4y^{2} - 6(3(x+2)) = 0$$9x(x+2) + 4y^{2} - 18(x+2) = 0$$9x^{2} + 18x + 4y^{2} - 18x - 36 = 0$$9x^{2} + 4y^{2} = 36$Dividing by $36$, we get $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$, which represents an ellipse.

Class 11 Mathematics Straight Line Locus of Point

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For the variable $t$, the locus of the point of intersection of the lines $3tx - 2y + 6t = 0$ and $3x + 2ty - 6 = 0$ is

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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 $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$

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For the variable $t$, the locus of the point of intersection of the lines $3tx - 2y + 6t = 0$ and $3x + 2ty - 6 = 0$ is

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