JEE Advanced 2017 Paper 1 Mathematics: Paragraph 1

For the line $y=mx+c$ to touch $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ we need $c^2=a^2m^2+b^2$. With $a^2=3$, $b^2=4$, $m=2$: $c^2=12+4=16$, so $c=\pm4$, and only (iv) $y=2x+4$ qualifies. The contact point is $\left(-\frac{a^2m}{c},\frac{b^2}{c}\right)=\left(-\frac{6}{4},1\right)=\left(-\frac32,1\right)$, which is (P). Check: $\frac{9/4}{3}+\frac14=1$ and $2(-\frac32)+4=1$. The other lines fail: (i) needs $c^2=25$, (ii) $c^2=9$, (iii) $c^2=1$. Hence (II) $\to$ (iv) $\to$ (P). Option (II)(iv)(S) pairs the correct line with the hyperbola's contact point, and the remaining options choose a non-tangent line, so only the stated combination is fully correct. Therefore the answer is the combination of the ellipse, the line $y=2x+4$ and the point $(-3/2,1)$.

Test each curve against the line $y=2x+3$. Parabola $y^2=8x$: substituting gives $4x^2+12x+9=8x$, i.e. $4x^2+4x+9=0$, discriminant $16-144<0$, so it misses. Circle $x^2+y^2=5$: $5x^2+12x+4=0$ has discriminant $144-80>0$, so it cuts (not touches). Ellipse: $c^2=a^2m^2+b^2=16$ needed, but $c^2=9$; it cuts. Hyperbola $\frac{x^2}4-\frac{y^2}7=1$: tangent condition $c^2=a^2m^2-b^2=16-7=9$ holds. The contact point is $\left(-\frac{a^2m}{c},-\frac{b^2}{c}\right)=\left(-\frac83,-\frac73\right)$, which is (S); verify $\frac{64/9}{4}-\frac{49/9}{7}=\frac{16-7}{9}=1$. So (III) $\to$ (ii) $\to$ (S). Therefore the answer is the combination of the hyperbola, the line $y=2x+3$ and the point $(-8/3,-7/3)$. Checking the other three curves against this line first, as done above, rules out every alternative and confirms the unique fully correct code.

The normal at a point $(x_0,y_0)$ is parallel to the gradient of the curve, so it passes through the origin only if $(x_0,y_0)$ is parallel to the gradient there. Parabola $y^2=8x$ at $(\frac12,2)$: gradient $(-8,4)$, and $\frac12\cdot4-2\cdot(-8)\ne0$. Ellipse at $(-\frac32,1)$: gradient $\left(\frac{2x}3,\frac{y}2\right)=(-1,\frac12)$, cross product $-\frac32\cdot\frac12-1\cdot(-1)=\frac14\ne0$. Hyperbola at $(-\frac83,-\frac73)$: gradient $\left(\frac x2,-\frac{2y}{7}\right)=(-\frac43,\frac23)$, cross product $-\frac{16}9-\frac{28}9\ne0$. For the circle $x^2+y^2=5$ every normal is a radius, so it always passes through the origin. The circle is touched by (i) $y=2x+5$ at $(-2,1)$ (here $4+1=5$), which is (Q). So (IV) $\to$ (i) $\to$ (Q).