Question

The straight line y + x = 1 touches the parabola :

• A. $x^2+4y=0$
• B. $x^2-x+y=0$
• C. $4x^2-3x+y=0$
• D. $x^2-2x+2y=0$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $x^2+4y=0$
B $x^2-x+y=0$
C $4x^2-3x+y=0$

The straight line is $y+x=1$

$\Rightarrow$ $y=1-x$

$\left(A\right)$ $x^2+4y=0$

$\Rightarrow$ $x^2+4(1-x)=0$ [$y=1-x$ ]

$\Rightarrow$ $x^2-4x+4=0$

Now,

$D=b^2-4ac$

$\Rightarrow$ $D=(-4{)}^2-4(1\left)\right(4)$

$=16-16$

$=0$

$\left(B\right)$ $x^2-x+y=0$

$\Rightarrow$ $x^2-x+1-x=0$ [$y=1-x$ ]

$\Rightarrow$ $x^2-2x+1=0$

Now,

$D=b^2-4ac$

$\Rightarrow$ $D=(-2{)}^2-4(1\left)\right(1)$

$=4-4$

$=0$

$\left(C\right)$ $4x^2-3x+y=0$

$\Rightarrow$ $4x^2-3x+1-x=0$

$\Rightarrow$ $4x^2-4x+1=0$

Now,

$D=b^2-4ac$

$\Rightarrow$ $D=(-4{)}^2-4(4\left)\right(1)$

$=16-16$

$=0$

$\left(D\right)$ $x^2-2x+2y=0$

$\Rightarrow$ $x^2-2x+2(1-x)=0$

$\Rightarrow$ $x^2-2x+2-2x=0$

$\Rightarrow$ $x^2-4x+2=0$

Now,

$D=b^2-4ac$

$\Rightarrow$ $D=(-4{)}^2-4(1\left)\right(2)$

$=16-8$

$=8$

$\therefore$ We can see option $A,B$ and $C$ touches the parabola.