🛠️ JEE➗ Maths

If \(\frac{x}{ma}+\frac{y}{nb}=1\) touches the ellipse \(\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\), then

Q1

If \(\frac{x}{ma}+\frac{y}{nb}=1\) touches the ellipse \(\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\), then

a

\({m}^{2}=\frac{{n}^{2}}{{n}^{2}−1}\text{  }\mathrm{or}\text{  }{n}^{2}=\frac{{m}^{2}}{{m}^{2}−1}\)

b

\({m}^{2}=\frac{{n}^{2}}{{n}^{2}+1}\text{  }\mathrm{or}\text{  }{n}^{2}=\frac{{m}^{2}}{{m}^{2}+1}\)

c

\({m}^{2}=\frac{{n}^{2}+1}{{n}^{2}}\text{  }\mathrm{or}\text{  }{n}^{2}=\frac{{m}^{2}+1}{{m}^{2}}\)

d

\({m}^{2}=\frac{{n}^{2}-1}{{n}^{2}}\text{  }\mathrm{or}\text{  }{n}^{2}=\frac{{m}^{2}-1}{{m}^{2}}\)

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