a) Does a conic, y^2 = 12x have two tangents from the point (6, 9) ? Justify it with calculation. [3] b) The dot product of two non zero vectors gives a positive real number. Justify it with…

a) Tangents from the point $(6,9)$ to the conic $y^2=12x$

The given conic is a parabola $y^2=12x$. To determine whether two tangents can be drawn from the external point $(6,9)$, we use the condition for tangency to a parabola $y^2=4ax$ (here, $4a=12⟹a=3$).

The equation of the tangent to the parabola $y^2=4ax$ at a point $(x_1,y_1)$ is: $$y{y}_1=2a(x+x_1)$$ For our parabola $y^2=12x$, the tangent equation becomes: $$y{y}_1=6(x+x_1)$$ If the tangent passes through $(6,9)$, substitute $x=6$, $y=9$: $$9y_1=6(6+x_1)⟹3y_1=2(6+x_1)⟹3y_1=12+2x_1$$ Since $(x_1,y_1)$ lies on the parabola, it satisfies $y_1^2=12x_1$. Substitute $x_1=\frac{y_1^2}{12}$ into the tangent condition: $$3y_1=12+2\left(\frac{y_1^2}{12}\right)⟹3y_1=12+\frac{y_1^2}{6}$$ Multiply by 6 to clear the fraction: $$18y_1=72+y_1^2⟹y_1^2-18y_1+72=0$$ Solve the quadratic equation: $$y_1=\frac{18\pm \sqrt{(-18{)}^2-4\cdot 1\cdot 72}}{2}=\frac{18\pm \sqrt{324-288}}{2}=\frac{18\pm \sqrt{36}}{2}=\frac{18\pm 6}{2}$$ Thus, $y_1=12$ or $y_1=6$. Corresponding $x_1$ values:

• For $y_1=12$: $x_1=\frac{12^2}{12}=12$
• For $y_1=6$: $x_1=\frac{6^2}{12}=3$

Since there are two distinct points of tangency $(12,12)$ and $(3,6)$, two distinct tangents can be drawn from $(6,9)$ to the parabola.

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b) Dot product of two non-zero vectors gives a positive real number

The dot product (or scalar product) of two non-zero vectors $\vec{a}$ and $\vec{b}$ is defined as: $$\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\cos \theta$$ where $\theta$ is the angle between them.

Example: Let $\vec{a}=(1,2,3)$ and $\vec{b}=(4,5,6)$. Their dot product is: $$\vec{a}\cdot \vec{b}=(1)(4)+(2)(5)+(3)(6)=4+10+18=32$$ Here, $32>0$ because the angle $\theta$ between $\vec{a}$ and $\vec{b}$ is acute ($\cos \theta >0$).

Justification:

• If $\theta <90^{\circ }$, $\cos \theta >0⟹\vec{a}\cdot \vec{b}>0$.
• If $\theta =0^{\circ }$, $\cos \theta =1⟹\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|>0$.
• If $\theta =90^{\circ }$, $\cos \theta =0⟹\vec{a}\cdot \vec{b}=0$ (orthogonal vectors).
• If $\theta >90^{\circ }$, $\cos \theta <0⟹\vec{a}\cdot \vec{b}<0$.

Thus, for non-zero vectors, the dot product is positive only when the angle between them is acute. The example above satisfies this condition.