Mathematics Test - 2

PSQ is focal chord 

Let SQ = x

∴ Semi latus rectum, 4 = 12x/6+x ⇒ x = 3.

Let coordinates of point of contact be (x₁, y₁). Equation of tangent is xx₁ + yy₁ = a₂

Comparing it with y = x + c, 

x₁ = – a²/c, y₁ = a²/c

The given equation of parabola is 

\(\mathrm{y}=2 \mathrm{x}^{2}+\mathrm{x} \Rightarrow \mathrm{x}^{2}+\frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{2}\)

\(\Rightarrow \mathrm{x}^{2}+2 \times \frac{\mathrm{x}}{2} \times \frac{1}{2}+\frac{1}{4}=\frac{\mathrm{y}}{2}+\frac{1}{4}\)

\(\Rightarrow\left(x+\frac{1}{2}\right)^{2}=\frac{1}{2}\left(y+\frac{1}{8}\right)\)

is of the form \(\mathrm{X}^{2}=\frac{1}{2} \mathrm{Y}\ldots(1) \quad\) where \(\mathrm{A}=\frac{1}{8}\)

Focus of (1) is \(\left(0, \frac{1}{8}\right)\)

where \(\mathrm{X}=0, \mathrm{Y}=\frac{1}{8}\)

\(\Rightarrow \mathrm{x}+\frac{1}{4}=0\) and \(\mathrm{y}+\frac{1}{8}=\frac{1}{8}\)

\(\Rightarrow \mathrm{x}=\frac{-1}{4}\) and \(\mathrm{y}=0\)

\(\therefore\) focus of given parabola is \(\left(\frac{-1}{4}, 0\right)\)

Assume \(\theta=90^{\circ} \ldots \ldots \quad\{\because 0 \leq \theta \leq \pi / 2\}\)

\(\sin \theta+2 \cos \theta=1\)

\(\Rightarrow \sin 90^{\circ}+2 \cos 90^{\circ}=1\)

\(\Rightarrow 1=1\)

Now, put \(\theta=90^{\circ}\) in \(2 \sin \theta-\cos \theta\)

\(2 \sin \theta-\cos \theta\)

\(\Rightarrow 2 \sin 90^{\circ}-\cos 90^{\circ}\)

\(\Rightarrow 2(1)-0\)

\(\Rightarrow 2\)