The line y=x+5 touches | vedclass

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(A, B, C) The line is $y=x+5$, so $m=1$ and $c=5$.$(A)$ For parabola $y^2=4ax$, the condition for tangency is $c=\frac{a}{m}$. Here $4a=20 \Rightarrow

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the hyperbola $\frac{x^2}{29}-\frac{y^2}{4}=1$

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(A, B, C) The line is $y=x+5$, so $m=1$ and $c=5$.$(A)$ For parabola $y^2=4ax$, the condition for tangency is $c=\frac{a}{m}$. Here $4a=20 \Rightarrow a=5$. Thus $c=\frac{5}{1}=5$. The line is a tangent.$(B)$ For ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the condition for tangency is $c^2=a^2m^2+b^2$. Here $a^2=16, b^2=9$. Thus $c^2=5^2=25$ and $a^2m^2+b^2=16(1)^2+9=25$. The line is a tangent.$(C)$ For hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, the condition for tangency is $c^2=a^2m^2-b^2$. Here $a^2=29, b^2=4$. Thus $c^2=5^2=25$ and $a^2m^2-b^2=29(1)^2-4=25$. The line is a tangent.$(D)$ For circle $x^2+y^2=r^2$, the condition for tangency is $c^2=r^2(1+m^2)$. Here $r^2=25, m=1$. Thus $c^2=25$ and $r^2(1+m^2)=25(1+1)=50$. Since $25 
eq 50$, the line is not a tangent.Therefore, options $(A)$, $(B)$, and $(C)$ are correct.

The line $y=x+5$ touches

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