JEE MAIN - Mathematics (2018 - 16th April Morning Slot - No. 21)
The locus of the point of intersection of the lines, $$\sqrt 2 x - y + 4\sqrt 2 k = 0$$ and $$\sqrt 2 k\,x + k\,y - 4\sqrt 2 = 0$$ (k is any non-zero real parameter), is :
an ellipse whose eccentricity is $${1 \over {\sqrt 3 }}.$$
an ellipse with length of its major axis $$8\sqrt 2 .$$
a hyperbola whose eccentricity is $$\sqrt 3 .$$
a hyperbola with length of its transverse axis $$8\sqrt 2 .$$
Explanation
Here, lines are :
$$\sqrt 2 x$$ $$-$$ y + 4$$\sqrt 2 k$$ = 0
$$ \Rightarrow $$$$\,\,\,$$ $$\sqrt 2 x + 4\sqrt 2 k = y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....$$(i)
and $$\sqrt 2 kx + ky - 4\sqrt 2 = 0\,\,\,\,\,...\left( {ii} \right)$$
Put the value of y from (i) in (ii) we get;
$$ \Rightarrow $$2$$\sqrt 2 $$kx + 4$$\sqrt 2 $$(k2 $$-$$ 1) = 0
$$ \Rightarrow $$ x = $${{2\left( {1 - {k^2}} \right)} \over k}$$, y = $${{2\sqrt 2 \left( {1 + {k^2}} \right)} \over k}$$
$$\therefore\,\,\,$$ $${\left( {{y \over {4\sqrt 2 }}} \right)^2} - {\left( {{x \over 4}} \right)^2} = 1$$
$$\therefore\,\,\,$$ length of transverse axis
2a = 2 $$ \times $$ 4$${\sqrt 2 }$$ = 8$${\sqrt 2 }$$
Hence, the locus is a hyperbola with length of its transverse axis equal to 8$${\sqrt 2 }$$
$$\sqrt 2 x$$ $$-$$ y + 4$$\sqrt 2 k$$ = 0
$$ \Rightarrow $$$$\,\,\,$$ $$\sqrt 2 x + 4\sqrt 2 k = y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....$$(i)
and $$\sqrt 2 kx + ky - 4\sqrt 2 = 0\,\,\,\,\,...\left( {ii} \right)$$
Put the value of y from (i) in (ii) we get;
$$ \Rightarrow $$2$$\sqrt 2 $$kx + 4$$\sqrt 2 $$(k2 $$-$$ 1) = 0
$$ \Rightarrow $$ x = $${{2\left( {1 - {k^2}} \right)} \over k}$$, y = $${{2\sqrt 2 \left( {1 + {k^2}} \right)} \over k}$$
$$\therefore\,\,\,$$ $${\left( {{y \over {4\sqrt 2 }}} \right)^2} - {\left( {{x \over 4}} \right)^2} = 1$$
$$\therefore\,\,\,$$ length of transverse axis
2a = 2 $$ \times $$ 4$${\sqrt 2 }$$ = 8$${\sqrt 2 }$$
Hence, the locus is a hyperbola with length of its transverse axis equal to 8$${\sqrt 2 }$$
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