and \[2{{x}^{2}}-4{{y}^{2}}=9\]is :
JEE Main Online Paper ( Held On 25 April 2013 )
A) \[x-y=\frac{3}{2}\]
B) \[x+y=1\]
C) \[x+y=\frac{9}{2}\]
D) \[x-y=1\]
Correct Answer: A
Solution :
\[{{x}^{2}}-6y\] ?(i) \[2{{x}^{2}}-4{{y}^{2}}=9\]Consider the line, \[x-y=\frac{3}{2}\] On solving (i) and (iii), we get only\[x=3,y=\frac{3}{2}\] Hence \[\left( 3,\frac{3}{2} \right)\] is the point of contact of conic (i), and line (iii) On solving (ii) and (iii), we get only x = 3,\[y=\frac{3}{2}\] Hence \[\left( 3,\frac{3}{2} \right)\]is also the point of contact of conic (ii) and line (iii). Hence line (iii) is the common tangent to both the given conies.
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